. 


AUAMS'S    MEXSljRATION, 


MENSlIxATlON. 


MECHANICAL   POWERS,] 


MACHINERY: 


BUNG 


A   SEQUEL 


KEVISED  EDITION  0?  :-.nyc?$  NEW  AlUTHilL  : 


/.  MISSIONED   FOB   THE   USE   OF    SCHOOLS   AND   ACADEMIES. 


NE\V*YORK: 
PUBLISHED    3Y    ROBERT    3. 

BOSTON  :    PHILLIPS,   SAMPAN   «fe   CO 

KEEIXfE,  N,  H. :  J.  W-  PBlfiNTISS  &  CO. 


I 


Mrs.  John  R.  Connolly 
1160  Lake  Road  West 
Williamson,  New  York 


ADAMS'S    M 


MENSURATION, 


MECHANICAL   POWEKS, 


AND 


MACHINERY. 


THE   PRINCIPLES   OF   MENSURATION   ANALYTICALLY  EXPLAINED,  AND 
PRACTICALLY  APPLIED  TO  THE  MEASUREMENT  OF  LINES,  SUPER- 
FICIES, AND  SOLIDS;  ALSO,  A  PHILOSOPHICAL  EXPLANATION 
OF  THE  SIMPLE  MECHANICAL  POWERS,  AND  THEIR 
APPLICATION  TO  MACHINERY. 


DESIGNED  FOR  THE  USE  OF  SCHOOLS  AND  ACADEMIES. 


NEW    YORK: 

ROBERT    B.    COLLINS. 
1850. 


V 


Entered  according  to  Act  of  Congress,  in  the  year  1849,  by 

J.    HOMER   FRENCH, 
In  the  Clerk's  Office  of  the  District  Court  of  the  District  of  New  Hampshire. 


Stereotyped    by 
HOBART  &  ROBBINS; 

NEW   ENGLAND   TYPE   AND    STEREOTYPE    FOUNDER*, 
BOSTON, 


PREFACE 


MORE  than  nineteen  twentieths  of  the  children  in  our  country  receive 
all  their  education  in  the  common  schools.  And  but  about  one  half  of 
the  number  who  attend  the  high  schools  and  academies,  ever  go  further 
in  a  course  of  mathematical  study  than  through  the  elements  of  Algebra 
and  Geometry.  Hence,  of  the  whole  number  of  scholars  in  all  the  schools 
in  our  country,  not  more  than  one  fortieth  ever  acquire  as  much  knowl- 
edge of  the  principles  of  Mensuration  as  is  actually  needed  for  the  every- 
day business  of  life. 

Now,  this  is  manifestly  wrong.  Every  person,  and  more  especially 
every  young  man,  should  possess  sufficient  knowledge  of  the  principles 
of  Mensuration  to  enable  him  to  transact  his  business  independent  of 
arbitrary  rules,  or  of  the  assistance  of  those  who,  having  been  more  for- 
tunate than  himself  in  acquiring  a  knowledge  of  these  principles,  render 
the  necessary  aid  only  for  a  stipulated  sum. 

But  wherein  lies  this  evil  ?  It  is  not  in  the  want  of  treatises  upon 
Mensuration ;  for  the  world  is  well  supplied,  upon  this  subject,  with  text- 
books abounding  in  mechanical  rules.  Neither  is  it  in  the  want  of  facil- 
ities for  acquiring  a  thorough  mathematical  knowledge ;  for  the  doors  of 
our  high  schools,  seminaries  and  colleges,  are  open  alike  to  all  who  may 
choose  to  enter.  The  rules  and  principles  presented  in  most  of  the  text- 
books heretofore  written  upon  this  department  of  education,  are  mere 
directions  for  the  performance  of  a  mechanical  process,  which,  if  followed, 
will  "  bring  the  answer." 

The  reason  generally  given  for  thus  presenting  them  is,  that  the  prin- 
ciples involved  cannot  be  understood  without  a  thorough  knowledge  of 
Geometry.  But  this  is  not  so.  A  great  majority,  nay,  nearly  all  of  the 
rules  and  principles  involved  in  Mensuration  as  applied  to  the  actual 
business  of  life,  admit  of  an  analysis  perfectly  comprehensible  by  the 
mere  arithmetician. 

The  evil  must  be,  then,  in  the  want  of  the  proper  kind  of  text-book  ;  — 
one  that  shall  give  the  why,  as  well  as  the  how  ;  one  that  shall  be  adapt- 
ed to  the  capacity  of  the  student  who  has  no  knowledge  of  mathematics 
beyond  Arithmetic. 

Such  a  work,  it  is  believed,  is  here  presented  to  the  public.  The  char- 
acteristics of  the  work  are  the  following  : 


IV  PREFACE. 

1.  It  is  an  analytical  rvorJc.    No  rule  or  principle  is  introduced  as 
mechanical,  that  admits  of  aii  analysis  intelligible  to  the  mind  of  the 
thorough  arithmetical  scholar.    After  the  number  of  rales  that  admit  of 
such  an  analysis  are  taken  from  the  whole  number  contained  in  the  work, 
it  leaves  but  a  very  small  number  of  arbitrary  mechanical  rales. 

2.  The  arrangement  is  natural  and  philosophical  The  subject  of  Weights 
and  Measures  is  first  considered,  for  the  reason  that  nothing  can  be  meas- 
ured without  reference  to  some  established  standard  of  weight  or  meas- 
ure.   The  Geometrical  Definitions,  necessary  to  be  understood  by  the 
pupil  in  pursuing  this  study,  are  next  introduced;  and  these  are  followed 
by  a  number  of  important  Geometrical  Problems.    The  Mensuration  of 
Lines  and  Superficies  is  then  presented,  and  next  in  order  follows  the 
Mensuration  of  Solids ;  care  being  taken  in  all  cases  to  present  the  vari- 
ous rules  in  their  most  natural  order.    The  Simple  Mechanical  Powers 
are  next  considered ;  and  the  work  closes  with  an  application  of  the 
Mechanical  Powers  to  machines,  and  an  examination  of  some  of  the 
important  principles  of  Machinery. 

4.  The  "  Topic  Method  "  of  questioning,  which  was  followed  in  "  Ad- 
ams's Book-keeping,"  having  been  received  with  much  favor,  is  adopted 
in  this  work.  This  method  points  out  something  for  the  pupil  to  do,  and 
it  also  requires  him  to  do  it. 

4.  In  the  analysis  of  the  various  principles,  and  in  the  examples  for 
practice,  care  has  been  taken  to  avoid  the  extremes  of  analysis  and  syn- 
thesis.   The  work  is  therefore  neither  so  obscure  as  to  be  unintelligible 
to  the  majority  of  pupils,  nor  so  puerile  as  to  leave  nothing  upon  which 
the  active  and  inquiring  mind  may  exercise  and  improve  itself. 

5.  The  analysis  of  many  of  the  rules  and  principles,  and  the  peculiar 
manner  in  which  the  subjects  generally  are  presented,  are  believed  to  be 
original.     The  Encyclopaedia  Britannica,  North  American  Keview,  the 
works  of  Dr.  Lardner,  Galloway,  Coulomb,  Rennie,  Willis,  and  Gregory, 
and  many  of  the  first  teachers,  machinists  and  mechanics,  in  New  Eng- 
land and  New  York,  have  been  consulted  in  the  preparation  of  the  work. 

The  work  contains  just  the  kind  of  information  required  by  the  mass 
of  people  throughout  the  country ;  and  it  is  confidently  hoped  that  its 
arrangement,  and  its  adaptation  to  the  best  and  most  approved  methods 
of  teaching,  together  with  the  importance  of  the  subject,  may  secure  for 
the  work  a  place  in  the  course  of  instruction  in  all  our  schools  and  acad- 
emies, though  it  be,  in  some  cases,  at  the  expense  of  some  of  the  higher 
but  less  important  branches. 


INDEX. 


WEIGHTS  AND  MEASURES. 

WEIGHTS. 

I.        Troy  Weight, 9 

II.         Apothecaries'  Weight, 10 

III.      .  Avoirdupois  or  Commercial  Weight, 10 

MEASURES  OP  EXTENSION. 

I.  1.  Linear  Measure, 11 

2.  Cloth  Measure, 12 

3.  Linear  Chain  Measure, 12 

4.  Duodecimal  Linear  Measure, 12 

5.  Miscellaneous  Linear  Measures, 12 

II.  1.  Square  Measure, 13 

2.  Square  Chain  Measure 14 

3.  Artificers'  Superficial  Measure, 14 

HI.         Cubic  or  Solid  Measure, 14 

MEASURES  OP  CAPACITY. 

I.        Wine  Measure, 15 

II.         Beer  Measure, 16 

HI.         Dry  Measure, 17 

Standard  Road  Measures, 18 

DEFINITIONS. 

GEOMETRICAL  DEFINITIONS. 

Lines  and  Angles, 21 

Plane  Figures, 23 

Rectilinear  Plane  Figures 23 

Curvilinear  and  Mixtilinear  Plane  Figures 27 

Solids  or  Bodies, ' 30 

PEACTICAL  GEOMETRY. 

GEOMETRICAL  PROBLEMS. 

Prob.    1.  To  draw  a  line  parallel  to  a  given  line, 35 

Prob.    2.  To  bisect  a  given  line, 36 

Prob.    3.  To  bisect  a  given  curve, 36 

Prob.    4.  To  bisect  a  given  angle,     36 

Prob.    5.  To  erect  a  perpendicular  on  the  middle  of  a  given  line, 37 

Prob.    6.  To  erect  a  perpendicular  on  any  given  point  in  a  line, 37 

Prob.    7.  From  any  point  without  a  given  line  to  draw  a  perpendicular  to  the  line,  .  37 

Prob.    8.  To  describe  a  circle  which  shall  pass  through  any  three  given  points  not 

in  a  right  line, 33 

Prob.    9.  To  find  the  center  of  a  circle, 33 

Prob.  10.  To  find  the  center  of  a  circle  of  which  an  arc  only  is  given, 33 

Prpb.  11.  To  draw  a  curve  parallel  to  a  given  curve, 39 

Prob.  12.  To  describe  a  right-angled  triangle,  the  base  and  perpendicular  being  given, .  39 

Prob.  13.  To  describe  an  equilateral  triangle, 39 


VI  INDEX. 

Prob.  14.  To  describe  a  triangle,  the  three  sides  being  given, 39 

Prob.  1 5.  To  describe  a  right-angled  triangle,  the  hypotenuse  and  one  side  being  given, .  40 

Prob.  16.  To  make  an  angle  equal  to  a  given  angle, 40 

Prob.  17.  To  describe  a  triangle,  two  sides  and  the  angle  which  they  contain  being 

given, 40 

Prob.  18.  To  describe  a  square, 40 

Prob.  19.  To  describe  a  rectangle, 41 

Prob.  20.  To  describe  a  rhombus, 41 

Prob.  21.  To  inscribe  a  triangle  in  a  circle 41 

Prob.  22.  To  inscribe  a  square  in  a  circle, 4,4 

Prob.  23.  To  inscribe  a  pentagon  in  a  circle, 42 

Prob.  24.  To  inscribe  a  hexagon  in  a  circle, 42 

Prob.  25.  To  inscribe  an  octagon  in  a  circle, 4-2 

Prob.  26.  To  inscribe  a  decagon  in  a  circle, 43 

Prob.  27.  To  inscribe  a  dodecagon  in  a  circle, 43 

Prob.  28.  To  inscribe  any  regular  polygon  in  a  circle, 43 

Prob.  29.  To  describe  any  regular  polygon, 4J 

Prob.  30.  To  circumscribe  a  regular  polygon  about  a  circle, 44 

Prob.  31.  To  circumscribe  a  circle  about  a  regular  polygon, 44 

Prob.  32.  To  inscribe  a  circle  in  a  regular  polygon, 44 

Prob.  33.  To  inscribe  a  circle  in  a  triangle, 44 

Prob.  34.  To  construct  solids, 45 

MENSURATION  OF  LINES  AND  SUPERFICIES. 

The  length  and  breadth  of  a  square  or  rectangle  being  given,  to  find  the  area,  .  .  .  .  47 
The  area  and  one  side  of  a  square  or  rectangle  being  given,  to  find  the  other  side,  ...  47 
The  base  and  perpendicular  of  a  right-angled  triangle  being  given,  to  find  the  hypot- 

The  hypotenuse  and  one  les  of  a  righVangledViangie  being  given,  to  find  the  other  leg,  .  49 

The  sum  and  difference  of  two  numbers  being  given,  to  find  the  numbers 50 

The  sum  of  two  numbers  and  the  difference  of  their  squares  being  given,  to  find  the 

numbers, ,•  •  °" 

One  side  of  a  right-angled  triangle,  and  the  sum  of  the  hypotenuse  and  the  other  side 

being  given,  to  find  the  hypotenuse  and  the  other  side, 51 

The  relation  of  the  three  sides  of  a  triangle  to  each  other,  applied  to  the  measurement 

of  distances, °2 

To  fimlthe  area  of  a  right-angled  triangle, «« 

To  find  the  area  of  an  equilateral  and  of  an  isosceles  triangle,  o5 

To  find  the  area  of  any  triangle, »b 

To  find  the  area  of  a  rhombus  and  of  a  rhomboid, 57 

To  find  the  area  of  a  trapezoid, 5° 

To  find  the  area  of  a  trapezium, »3 

Similar  Rectilinear  Fieures, 5x 

To  find  the  area  of  any  regular  polygon °y 

Table  of  Regular  Polygons °l 

To  find  the  area  of  any  rectilinear  figure  or  polygon, ^ 

Board  or  Lumber  Measure, bA 

The  diameter  of  a  circle  being  given,  to  find  the  circumference, b4 

The  circumference  of  a  circle  being  given,  to  find  the  diameter, 66 

The  number  of  degrees  in  a  circular  arc,  and  the  radius  of  the  circle  being  given,  to 

find  the  length  of  the  arc, 66 

To  find  the  area  of  a  circle °4 

To  find  the  area  of  a  circle  when  the  diameter  only  is  given, 67 

To  find  the  diameter  of  a  circle,  the  area  being  given, od 

To  find  the  area  of  a  semicircle,  a  quadrant,  and  a  sextant, b» 

To  find  the  area  of  a  sector,  the  radius  and  arc  being  given, b9 

To  find  the  area  of  a  sector,  the  radius  and  the  angle  at  the  center  being  given,  ....  70 
To  find  the  side  of  a  square  which  shall  contain  an  area  equal  to  a  given  circle,  ...  70 

To  find  the  side  of  an  equilateral  triangle  inscribed  in  a  given  circle, 71 

To  find  the  side  of  a  square  inscribed  in  a  given  circle, '2 

To  find  the  side  of  an  octagon  inscribed  in  a  given  circle, « 

To  find  the  area  of  an  ellipse, I'',''.''  ;,-'  '  '  '  ~ , 

To  find  the  diameter  of  a  circle  whose  area  shall  be  equal  to  that  of  a  given  ellipse  .  .  74 
To  find  the  area  of  the  space  contained  between  the  arcs  of  four  equal  adjacent  circles,  .  74 
To  find  the  area  of  the  space  contained  between  the  arcs  of  three  equal  adjacent  circles,  .  75 

To  find  the  area  of  a  circular  ring, £° 

Similar  Curvilinear  Figures, '° 

Practical  Examples  in  the  Mensuration  of  Lines  and  Superficies, ,....// 


INDEX.  VH 


MENSURATION  OF  SOLIDS, 

To  find  the  cubic  contents  of  a  prism,  cube,  parallelepiped,  cylinder,  or  cylindroid,  .  80 

To  find  the  cubic  contents  of  a  pyramid  or  a  cone, 81 

To  find  the  night  of  a  pyramid  or  a  cone,  of  which  a  given  frustrum  is  a  part,  ....  82 

To  find  the  solidity  of  a  frustrum  of  a  pyramid  or  a  cone, 83 

To  find  the  superficies  and  the  solidity  of  the  regular  solids, 84 

Table  of  Regular  Solids, 84 


To  find  the  solidity  of  any  irregular  solid, 
To  find  the  area  of  a  sphere, 


To  find  the  solidity  of  a  sphere, 

Gauging, 87 

Timber  Measure, 88 

To  find  the  contents  of  a  four-sided  stick  of  timber  which  tapera  upon  two  opposite 

sides  only, 88 

To  find  the  contents  of  a  stick  of  timber  which  tapers  uniformly  upon  all  sides,   ...  89 

To  find  the  number  of  cubic  feet  of  timber  any  log  will  make  when  hewn  square,   .   .  89 

To  find  the  number  of  feet  of  boards  that  can  be  sawn  from  any  log  of  a  given  diameter,  .  90 

To  find  how  many  bushels  of  grain  may  be  put  into  a  bin  of  a  given  size, 92 

Table  for  Boxes  or  Measures,  Dry  Measure, 93 

To  find  the  side  of  the  greatest  cube  that  can  be  cut  from  any  sphere, 93 

To  find  the  weight  of  lead  and  iron  balls, 94 

Practical  Examples  in  the  Mensuration  of  Solids, 94 

MECHANICAL  POWERS, 

The  Lever, 98 

The  Wheel  and  Axle, 101 

The  Pulley, 103 

Smeaton's  Pulley, 105 

The  Inclined  Plane, 106 

The  Wedge, 107 

The  Screw, 108 

Friction, 110 

The  Friction  of  Sliding  Bodies, 110 

The  Friction  of  Rolling  Bodies, 110 

The  Friction  of  the  Axles  of  Wheels Ill 

General  Remarks  upon  the  Mechanical  Powers, 112 

MACHINERY. 

Methods  of  Transmitting  Motion, 113 

Spur,  Crown,  and  Beveled  or  Conical  Wheels, 114 

The  Universal  Joint 115 

Teeth  of  Wheels, 115 

Horse  Power,     116 

Levers  and  Weighing  Machines 117 

Compound  Lever, 117 

The  Balance, 118 

The  Steelyard, 1 18 

The  Bent  Lever  Balance, 119 

Wheel  Work 119 

White's  Pulley, 121 

The  Crane, 122 

Hunter's  Screw, 123 

The  Endless  Screw, 123 

Pumps, 124 

The  Hydrostatic  Press,    ...    • 125 

The  Tread-Mill, 126 

Water  Wheels, 127 

The  Overshot  Wheel 127 

The  Undershot  Wheel, 127 

The  Breast  Wheel, 127 


ADAMS'S  ARITHMETICAL  SERIES. 

FOE  SCHOOLS  AND  ACADEMIES. 


L  — PRIMARY  ARITHMETIC,  OR  MENTAL  OPER- 
ATIONS IN  NUMBERS;  being  an  introduction  to  the 

REVISED  EDITION  OF  ADAMS'S  NEW  ARITHMETIC. 

II. —  ADAMS'S  NEW  ARITHMETIC,  REVISED  EDI- 
TION ;  in  which  the  principles  of  operating  by  numbers  are 
analytically  explained  and  synthetically  applied.  Illustrated 
by  copious  examples. 

III.  —  MENSURATION,    MECHANICAL    POWERS, 
AND   MACHINERY.     The  principles  of  Mensuration  ana- 
lytically explained,  and  practically  applied  to  the  measurement 
of  lines,  superficies  and  solids:  also,  a  philosophical  explanation 
of  the  simple  mechanical  powers,  and  their  application  to  ma- 
chinery. 

IV.  — BOOK-KEEPING;  containing  a  lucid  explanation 
of  the  common  method  of  BOOK-KEEPING  BY  SINGLE  ENTRY  ;  a 
new,  concise,  and  common-sense  method  of  Book-keeping,  for 
farmers,  mechanics,  retailers,  and  professional  men ;  methods 
of  keeping  books  by  figures  ;  short  methods  of  keeping  accounts 
in  a  limited  business ;   exercises  for  the  pupil ;  and  various 
forms  necessary  for  the  transaction  of  business.     Accompanied 
with  BLANK  BOOKS,  for  the  use  of  learners. 


ADVERTISEMENT. 

The  Primary  Arithmetic,  the  Treatise  on  Mensuration,  and 
the  Book-keeping,  have  been  mainly  prepared,  under  my  super- 
vision, by  Mr.  J.  HOMER  FRENCH,  of  New  York,  who  rendered 
important  assistance  in  revising  my  New  Arithmetic. 

From  my  knowledge  of  his  ability,  and  from  a  careful  ex- 
amination of  the  works,  I  can  confidently  say  they  meet  my 
approbation. 

DANIEL  ADAMS. 

Keene,  N.  H,  August,  1848. 


MENSURATION. 


WEIGHTS   AND  MEASURES. 

5T  1.  Measure  is  that  by  which  extent  or  dimension  is  as- 
certained, whether  it  be  length,  breadth,  thickness,  or  amount. 
The  process  by  which  the  extent  or  dimension  is  obtained 
is  called  Measuring,  which  consists  in  comparing  the  thing  to 
be  measured  with  some  conventional  standard. 

Weight  is  the  measure  of  the  force  by  which  any  body,  or  a 
given  portion  of  any  substance,  tends  or  gravitates  to  the  earth. 
The  process  by  which  this  measure  is  ascertained  is  called 
Weighing,  which  consists  in  comparing  the  thing  to  be  meas- 
ured with  some  conventional  standard. 

The  United  States  government,  after  various  unsuccessful 
attempts,  at  length  succeeded,  in  the  year  1834,  in  adopting  a 
uniform  standard  of  weights  and  measures,  for  the  use  of  the 
custom-houses,  and  the  other  branches  of  business  connected 
with  the  government.  In  the  following  tables  the  United 
States  standards  are  given. 

Weights. 

1.     TROY  WEIGHT. 

5f  2.  Troy  Weight  is  used  where  great  accuracy  is  re- 
quired, as  in  weighing  gold,  silver,  and  jewels.  The  denom- 
inations are  pounds,  ounces,  pennyweights  and  grains. 

TABLE. 

24  grains  (grs.)  make  1  pennyweight,  marked  pwt. 
20  pwts.  "       1  ounce,  "  oz. 

12  oz.  "       1  pound,  Ib. 

IT  1.    Measure.     Measuring.     Weight.    Weighing.     U.  S.  government 
standard  weights  and  measures. 
V  2.    Troy  Weight.    Denominations.    Table. 


*%  'S  * 

10  WEIGHTS  AND  MEASURES.  H  3-5. 

IT  3.  The  U.  S.  standard  unit  of  weight  is  the  Troy 
pound  of  the  mint,  which  is  the  same  as  the  Imperial  standard 
pound  of  Great  Britain. 

A  cubic  inch  of  distilled  water  in  a  vacuum,  weighed  by 
brass  weights,  also  in  a  vacuum,  at  a  temperature  of  62°  of 
Fahrenheit's  thermometer,  is  equal  to  252' 724  grains,  of  which 
the  standard  Troy  pound  contains  5760.  Consequently,  a  cubic 
inch  of  distilled  water  is  ^ffflfifa  of  a  standard  Troy  pound. 
Hence,  if  the  standard  Troy  pound  be  lost,  destroyed,  defaced, 
or  otherwise  injured,  it  may  be  restored  of  the  same  weight,  by 
making  a  new  standard,  determined  according  to  this  ratio. 

II.     APOTHECARIES'  WEIGHT. 

IT  4.  For  the  use  of  apothecaries  and  physicians,  in  com- 
pounding medicines,  the  Troy  ounce  is  divided  into  drams, 
the  drams  into  scruples,  and  the  scruples  into  grains. 

TABLE. 

20  grains  (grs.)  make  1  scruple,  marked  & 
SB  "  1  dram,  "  £ 

85  "1  ounce,  "  % 

12%  "1  pound,  "  Sb. 

Medicines  are  bought  and  sold  by  avoirdupois  weight. 

III.     AVOIRDUPOIS  or  COMMERCIAL  WEIGHT. 

^T  5.  Avoirdupois  Weight  (also  called  Commercial  Weight) 
is  employed  in  all  the  ordinary  purposes  of  weighing.  The 
denominations  are  tons,  pounds,  ounces,  and  drams. 

TABLE. 

16  drams  (drs.)  make  1  ounce,  marked  oz. 

16  oz.  "      1  pound,       "         Ib. 

2000  Ibs.  "      1  ton,  «        T. 

NOTE.  In  the  U.  S.  custom-house  operations,  hi  invoices  of  English  goods, 
and  of  coal  from  the  Pennsylvania  mines,  — 

23  Ibs.  make  1  quarter,  marked       qr. 

4  qrs.  =  ll2  Ihs.        "     1  hundred  weight,        "          cwt. 

20cwt.  =  2240  Ibs.      "      1  ton,  T. 

IT  3.  U.  S.  standard  unit  of  weight.    How  determined. 

IT  4.  Apothecaries'  weight.     Table. 

IT  5,  Avoirdupois  or  commercial  weight.  Denominations.  Table. 
Note. 


^  • 

If  6-8.  WEIGHTS  AND  MEASURES.  11 

But  in  selling  coal  in  cities,  and  in  other  transactions,  unless  otherwise  stipu- 
lated, 2000  Ibs,  are  called  a  ton. 

The  ton  of  2240  Ibs.  is  sometimes  called  the  "long  ton,"  and  the  ton  of 
2000  Ibs.  the  "short  ton," 

5T  6.  The  U.  S.  avoirdupois  pound  is  determined  from  the 
standard  Troy  pound,  and  contains  7000  Troy  grains,  the 
Troy  pound  containing  5760. 

The  Troy  pound  is  f  £§§  =  |^|-,  or  nearly  -if,  of  an  avoir- 
dupois pound.  But  the  Troy  ounce  contains  (-S-if-0- =)  480 
grs.,  and  the  avoirdupois  ounce  (^f  |p  =)  437'5  grs.  Troy. 
Therefore,  the  Troy  ounce  is  greater  than  the  avoirdupois 
ounce  in  the  ratio  of  480  to  437'5  =  4SOO  to  4735=192 
to  175. 

5T  7.  The  standard  pound  of  the  State  of  New  York  is  the 
pound  avoirdupois,  which  is  defined,  by  declaring  that  a  cubic 
foot  of  pure  water,  at  its  maximum  density,  (39'83°  Fahren- 
heit,) weighs  62'5  pounds,  or  1000  ounces,  using  brass  weights, 
at  the  mean  pressure  of  the  atmosphere  at  the  level  of  the  sea, 
(i.  e.,  the  barometer  being  at  30  inches.)  Therefore,  the 
standard  pound  of  the  State  of  N.  Y.  is  the  weight  of  27'64S 
cubic  inches  of  distilled  water,  weighed  in  air,  the  temperature 
being  39'83°  Fahrenheit,  and  the  barometer  at  30  inches. 


Measures  of  Extension. 

I.     1.  LINEAR  MEASURE. 

^T  8.  Linear  Measure  (also  called  Long  Measure)  is  the 
measure  of  lines ;  it  is  used  when  only  one  dimension  is  con- 
sidered, which  may  be  length,  breadth,  or  thickness. 

The  usual  dimensions  are  miles,  furlongs,  rods,  yards,  feet, 
and  inches. 

TABLE. 

12    inches  (in.)                             make  1  foot,       marked  ft. 

3    ft.                                               "      1  yard,          "  yd. 

5 1  =  5'5  yds.,  or  16  J  =  16<5  ft,    "      1  rod,            "  rd. 

40    rds.                                              "      1  furlong,      "  fur. 

8    fur.,  or  320  rds.,                          "      1  mile,           «  mi. 

IT  6.  U.  S.  avoirdupois  pound.  Ratio  of  the  Troy  pound  to  the  avoirdu- 
pois pound.  — of  the  Troy  ounce  to  the  avoirdupois  ounce. 

IT  7.  N.  Y.  standard  pound.  Maximum  density  and  mean  pressure  of  the 
atmosphere. 

IT  8.    Linear  measure.    Denominations.    Table. 


WEIGHTS  AND  MEASURES.  If  9-12. 


2.     CLOTH  MEASURE. 

IT  O.  Cloth  Measure  is  used  in  measuring  cloth  and  other 
goods  sold  by  the  yard  in  length,  without  regard  to  width. 

'  TABLE. 

2  J  =  2'25  inches  (in.)  make  1  nail,        marked  na. 
4  na.,  or  9  in.,  "      1  quarter,       "         qr. 

4  qrs.,  or  36  in.,  "      1  yard,  "        yd. 

3.    LINEAR  CHAIN  MEASURE. 

IT  1O.  The  Surveyor's,  or  what  is  called  Gunter's  Chain, 
is  generally  used  in  measuring  long  distances,  and  in  surveying 
land. 

It  is  4  rods,  or  66  feet,  in  length,  and  consists  of  100  links. 

TABLE. 

7'92  inches  (in.)  make  1  link,    marked  1. 

25       1.  «      1  rod,         "       rd. 

4       rds.,  66  ft.,  or  100  1.,     «      1  chain,      "        C. 
80       C.  «      1  mile,        "       mi. 

4.     DUODECIMAL  LINEAR  MEASURE. 

IT  1  1  .  Duodecimals  are  fractions  of  a  foot.  The  denom- 
inations are  fourths,  thirds,  seconds,  primes  or  inches,  and  feet 

TABLE. 

12  fourths  (  ""  )  make  1  third,  marked  "' 

12  '"  «       1  second, 

12  "  "      1  prime,  or  inch,       "        ' 

12  '  "1  foot,  «        ft. 

NOTE.  The  marks,  ',  ",  '",  "",  &c.,  which  distinguish  the  different  parts, 
are  called  the  indices  of  the  parts  or  denominations.  [See  Adams's  Revised 
Arithmetic,  ITU  203  and  204.] 

IT  13.    5.  MISCELLANEOUS  LINEAR  MEASURES. 

6  points  make  1  line,  )  used  in  measuring  the  length  of  the 
12  lines  "  1  inch,  \  rods  of  clock  pendulums. 


4  inches     «      1  hand,  j  j^fj11   ™*SU™S    *e    hight   of 


IT    9.  Cloth  measure.     Table. 

IT  10.  Linear  chain  measure.     Gunter's  chain.    Table. 

IT  11.  Duodecimal  linear  measure.    Denominations.    Table.    Note. 

T  19*  Miscellaneous  linear  measures. 


IF  13-15.  WEIGHTS  AND  MEASURES.  13 

6  feet     make  1  fathom,  used  in  measuring  depths  at  sea. 
18  inches     "      1  cubit. 
21<SS8  in.   "      1  sacred  cubit. 

69£  common  miles  make  1  degree,  or0,  j  on  the  equatorial  cir- 

'  (  cum.  of  the  earth. 

3  geographical  miles  make  1  league,  (L.)  \  "^  in  mef surin£ 

'  v     '  I  distances  at  sea. 

60  geographical  miles  make  1  degree  of  latitude. 

IT  13.  The  U.  S.  standard  of  measures  of  extension, 
whether  linear,  superficial,  or  solid,  is  the  yard  of  3  feet,  or  36 
inches,  and  is  the  same  as  the  Imperial  standard  yard  of  Great 
Britain.  The  standard  yard  is  made  of  brass,  and  is  laid  off 
from  a  scale  made  by  Troughton,  (a  celebrated  English  artist,) 
the  brass  being  at  the  temperature  of  62°  Fahrenheit's  ther- 
mometer. 

The  standard  yard,  when  compared  with  the  length  of  the 
rod  of  a  pendulum  vibrating  seconds  of  mean  time  in  the  lati- 
tude of  London,  in  a  vacuum  at  the  level  of  the  sea,  is  found 
to  be  in  the  ratio  of  36  inches  to  394393.  Hence,  if  the 
standard  yard  be  lost  or  destroyed,  it  may  be  restored,  by 
making  it  $f£$$$  of  the  length  of  the  rod  of  a  pendulum 
vibrating  seconds  under  the  above  described  circumstances. 

1T  14.  The  standard  yard  of  the  State  of  N.  Y.  is  a  brass 
rod,  which  bears  to  the  length  of  the  rod  of  a  pendulum  vibrating 
seconds  in  a  vacuum,  in  Columbia  College,  the  relation  of 
1000000  to  1086141,  the  brass  being  at  32°  Fahrenheit. 

II.     1.  SQUARE  MEASURE. 

^T  15.  Sgiiare  Measure  is  used  in  measuring  all  things 
wherein  length  and  breadth  are  considered;  such  as  land, 
flooring,  roofing,  painting,  plastering,  &c. 

The  denominations  are  miles,  acres,  roods,  rods  or  poles, 
yards,  feet,  and  inches. 

TABLE. 

144    square  inches  (sq.  in.)  make  1  square  foot,  marked  sq.  ft. 
9    sq.  ft.  "      1  square  yard,     "      sq.  yd. 

IT  13.  U.  S.  standard  of  measures  of  extension.  Material  of  standard 
yard.  Temperature.  Comparison  of  standard  yard  with  the  length  of  the  rod 
of  a  pendulum.  Standard  yard,  if  lost,  may  be  restored. 

IT  14.    N.  Y.  standard  yard. 

TT  15.    Square  measure.    Denominations.    Table. 
2 


!£  WEIGHTS  AND  MEASURES.  H  16-18. 

30J  =  30<25  sq.  yds.,  or  )  k  1     (  square  rod,  or  pole, 

*272J  =  272*25  sq.  ft.  1  maJ           1  marked  sq.  id.  or   P. 

40    sq.rds.  "      1  rood,             marked    R. 

4    R.,  or  160  sq.  ids.,  "      1  acre,                             A. 

640    £  "1  square  mile, 

2.    SQUARE  CHAIN  MEASURE. 

IT  16.     The   dimensions  of  land  are  generally  taken  by 
Gunter's  chain,  and  are  estimated  by  the  following 

TABLE. 

,     (     marked 
625  square  links  (sq.  1.)  make  1  square  rod,  or  pole,  |  ^  rd>  or  p 

jg  p  "1  square  chain,  marked       sq.  C. 

lOsq.C.  "      lacie,  A. 

640  A.,  or  6400  sq.  C.,    "      1  square  mile, 

3.    ARTIFICERS'  SUPERFICIAL  MEASURE. 

5T  17.     Artificers  estimate  their  work  as  follows : 

1.  Glazing,  and  stone-cutting,  by  the  square  foot. 

2.  Painting,  plastering,  paving,  ceiling,  and  paper-hanging, 
by  the  square  yard.  , 

3.  Flooring,  partitioning,  roofing,  slating  and  tiling,  by  tJ 
square  of  100  feet. 

4.  Brick-laying,  by  the  thousand  bricks ;  also,  by  the  squa 
yard,  and  the  square  of  100  feet. 

NOTE  1      In  estimating  the  painting  of  mouldings,  cornices,  &c.,  the  meas- 
uring line  is  carried  into  all  the  mouldings  and  cornices. 

NOTE  2.     In  estimating  brick-laying  by  the  square  yard,  or  by  the  square 
of  100  feet,  the  work  is  understood  to  be   li  ^^^j^^^JL 
it  be  of  any  other  thickness,  it  must  be  reduced  to  that  of  li  bricks,  be 
estimating  the  brick-laying. 

III.     CUBIC  OR  SOLID  MEASURE. 

IT  1 8.  Cubic  or  Solid  Measure  is  used  in  measuring  things 
that  have  length,  breadth,  and  thickness ;  such  as  timber,  wood, 
earth,  stone,  &c.  .  , 

The  denominations  are  cords,  tons,  yards,  feet,  and  i      .es. 

II  16.    Square  chain  measure.    Table. 

IT  17.    Artificers' superficial  measure.    Note  1.     Note  2. 

IT  18.  Cubic  or  solid  measure.  Denominations.  Table.  A  cubic  yard 
of  earth.  A  cubic  ton.  A  cord  of  wood.  A  cord  foot.  A  perch  of  stone,  or 
masonry.  Note  1.  Note  2. 


^  19.  WEIGHTS  AND  MEASURES.  15 

TABLE. 

1728  cubic  inches  (cu.  in.)  make   1  cubic  foot,    marked  cu.  ft. 

27  cu.  ft.                             "       1  cubic  yard,          "  cu.  yd. 

50  cu.  ft.  of  round  timber,  or    )    1  ,                           «  T 
40  cu.  ft.  of  hewn  timber,  "  ] 

42  cu.  ft                             "       1  ton  of  shipping,  "  T. 

f .                                    (  1  cord  foot,  or  )      u  /-«  f. 

16cu'ft                                  {l  foot  of  woodj  C'ft' 

8  G.  ft.,  or  128  cu,  ft.,      "       1  cord  of  wood,      «  C. 

24f  =  24'75  cu.  ft.            «       1  perch  of  stone,    "  Pch. 

A  cubic  yard  of  earth  is  called  a  load. 

A  cubic  ton  is  used  for  estimating  the  cartage  and  trans- 
portation of  timber.  A  ton  of  round  timber  is  such  a  quantity 
(about  50  feet)  as  will  make  40  feet  when  hewn  square. 

A  pile  of  wood  8  feet  long,  4  feet  wide,  and  4  feet  high,  con- 
tains 1  cord;  and  a  cord  foot  is  1  foot  in  length  of  such  a  pile. 

A  perch  of  stone  or  of  masonry  is  16 J  feet  long,  1|  feet 
wide,  and  1  foot  high.  If  any  wall  be  1J  feet  thick,  its  con- 
tents in  perches  will  be  equal  to  the  number  of  times  16J 
square  feet  are  contained  in  the  superficial  contents  of  the  wall 
expressed  in  feet.  If  the  wall  be  of  any  other  thickness,  the 
number  of  perches  it  contains  will  be  found  by  dividing  its 
cubic  contents  by  the  cubic  contents  of  a  perch. 

NOTE  1.  Joiners,  brick-layers,  and  masons  make  no  allowance  for  win- 
dows, doors,  &c.  Brick-layers  and  masons,  in  estimating  their  work  by  cubic 
measure,  make  no  allowance  for  the  corners  to  the  walls  of  houses,  cellars,  &c., 
but  estimate  their  work  by  the  girt,  that  is,  the  length  of  the  wall  on  the  out- 
side. 

NOTE  2.  Engineers,  in  making  estimates  for  excavations  and  embankments, 
take  the  dimensions  with  a  line,  or  measure,  divided  into  feet  and  decimals  of 
a  foot.  The  estimates  are  made  in  feet  and  decimals,  and  are  then,  reduced  to 
cubic  yards. 


Measures  of  Capacity. 

I.     WINE  MEASURE. 

IT  1O.     Wine  Measure  is  used  in  measuring  all  liquids 
except  ale,  beer  and  milk. 

The  denominations  are  gallons,  quarts,  pints,  and  gills. 

TT  19.    Wine  measure.    Denominations.    Table.     Note. 


WEIGHTS  AND  MEASURES.  1F  20-22. 

TABLE. 

4  gills  (gi.)  make  1  pint,      marked    pt. 
2  pts.  «      1  quart,         "         qt. 

4  qts.  «      1  gallon,        »       gal. 

NOTE.    The  following  denominations  are  also  sometimes  used  in  this  meas- 

31 J  gal.  make  1  barrel,  marked  bar. 

42    gal.  «       1  tierce,  «        tier. 

63    gal.,  or  2  bar.,     "      1  hogshead,  "       hhd. 

hhds.  «       1  plpe,  or  butt,       »  P. 

2    P.,  or  4  hhds.,    "      1  tun,  «          T. 

But  the  tierce,  hogshead,  puncheon,  pipe,  butt,  and  tun,  used  for  liquids- 
are  so  vague  and  variable  in  their  contents,  that  they  are  to  be  considered  rather 
as  the  names  of  casks  than  as  expressing  any  fixed  or  definite  measures 
buch  vessels  are  usually  gauged,  and  have  their  contents  marked  on  them. 

IT  SO.  The  U.  S.  standard  of  liquid  measure  is  the  old 
English  wine  gallon,  of  231  cubic  inches,  equal  to  8'339 
pounds  avoirdupois  of  distilled  water,  at  the  maximum  density 
39'83°  Fahrenheit*  the  barometer  at  30  inches. 

IF  21.  The  standard  of  liquid  measure  in  the  State  of  New 
York  is  the  wine  gallon,  which  the  legislature  have  defined  to 
be  equal  to  8  pounds  of  pure  water  at  its  maximum  density. 
Hence  the  N.  Y.  wine  gallon  contains  221' 184  cubic  inches. 

II.     BEER  MEASURE. 

1T  22.  Beer  Measure  is  used  in  measuring  beer,  ale,  and 
milk. 

The  denominations  are  hogsheads,  barrels,  gallons,  quarts, 
and  pints. 

TABLE. 

2  pints  (pts.)         make  1  quart,        marked     qt. 
4  qts.  «       1  gallon,  «         gal. 

36  gal.  «      1  barrel,  «        bar. 

54  gal,  or  1J  bar.,     "      1  hogshead,       "       hhd. 
The  gallon  Beer  Measure  contains  282  cubic  inches. 
Beer  Measure  is  retained  in  use  only  by  custom.     In  many 
places  its  use  is  entirely  discarded. 

IT  20.     U.  S.  standard  of  liquid  measure. 
IT  21.    N.  Y.  standard  of  liquid  measure. 

IT  22.    Beer  measure.    Denominations.    Table.     Beer  gallon.    Authority 
for  using  this  measure. 


1T  23-25.  WEIGHTS  AND  MEASURES.  17 


III.     DRY  MEASURE. 

IT  S3.     Dry  Measure  is  used  in  measuring  grain,  fruit, 

roots,  salt,  coal,  &c. 

The  denominations  are  bushels,  pecks,  auarts  and  pints. 

TABLE. 

2  pints  (pts.)  make  1  quart,    marked  qt. 

8  qts.  "      1  peck,  "  pk. 

4  pks.  "      1  bushel,       "      bu.,  or  bush. 

The  quarter  of  8  bushels  is  an  English  measure  for  grain. 
The  chaldron  of  36  bushels  is  sometimes  used  in  measuring 
charcoal. 

IT  24.     The  U.  S.  standard  of  dry  measure  is  the  British 

Winchester  bushel,  which  is  18J  inches  in  diameter,  and  8 
inches  deep,  and  contains  2150*4  cubic  inches,  equal  to  77'6274 
pounds  of  distilled  water,  at  the  maximum  density.  A  gallon 
dry  measure  contains  26S'8  cubic  inches. 

IT  25.  The  standard  bushel  of  the  State  of  N.  Y.  conta ins  80 
pounds,  or  2211'84  cubic  inches  of  pure  water  at  its  maximum 
density;  the  gallon  contains  10  pounds,  or  276'48  cubic 
inches. 

NOTE.  The  Imperial  gallon  of  Great  Britain,  for  all  liquids  and  dry  sub- 
stances, contains  277'274  cubic  inches,  or  10  pounds  avoirdupois  weight  of  dis- 
tilled water  weighed  in  air,  at  62°  Fahrenheit,  the  barometer  at  30  inches. 

The  Imperial  standard  bushel  contains  2218'192  cubic  inches,  or  80  pounds 
of  distilled  water,  weighed  in  the  manner  above  described. 

TF  23.    Dry  measure.    Denominations.     Table.     Quarter .    Chaldron. 
IF  24.    U.  S.  standard  of  dry  measure.    Dry  gallon. 
IT  25.    N.  Y.  standard  bushel.    N.  Y.  dry  gallon.    Imperial  gallon  and 
bushel  of  Great  Britain. 
2* 


18 


WEIGHTS  AND  MEASURES. 


IF  26. 


IF  26.     STANDARD   ROAD  MEASURES  OF  DIFFERENT 
COUNTRIES. 


Countries. 

Measures. 

Yards. 

English 
mile. 

629 

'357 

Corsica,  France,  French  Gui- 
ana,  Netherlands,  Modern 
Greece,  and  French  West 

Mile      

1093*633 

'621 

Russia,                               •     • 

Werst,  

1167 

'663 

Sicily,    

Mile,     

1405*555 

'798 

Ancient  Greece,    .... 
Ancient  Rome,      .     .     . 
Modern  Rome        .... 
Australia,    Canada,    Ceylon, 
Gibraltar,    Great    Britain, 
British  Guiana,  N.  Bruns- 
wick, Newfoundland,  Nova 
Scotia,  United  States,  and 
British  West  Indies,   .     . 
Italy,      
Tuscany                           . 

t  i 

u 
11 

11 

Old  common  mile,     . 
Mile,     

1611'666 
1614 
1628 

1760 
1766'333 

1808 

'915 
'917 
'925 

I'OOO 
1*003 
1'027 

Tuombardv 

a 

1813*432 

i'030 

Turkey,  

Barri,    

1826 

1*037 

Ancient  mile,  . 

1984 

1'127 

Bengal,  (India,)    .... 

League,      .... 

2000 

1*136 

Naples 

Mile,     

2018 

1*146 

Australia,    Canada,    (.Vvlon. 
Gibraltar,    Great    Britain, 
British  Guiana,  N.  Bruns- 
wick, Newfoundland,  Nova 
Scotia,  United  States,  anc 
British  West  Indies,   .     . 
Arabia,  
Ireland,  
Brazil,  Madeira,a«id  Portugal 
Ancient  Jewish,    .... 
Piedmont 

Geographical  mile,   . 
Mile,     

2025 
2148 
2210 
2253 
2432 
2640 

1'150 

'220 
'272 
'280 
'381 
'500 

2697 

'532 

Corsica,  France,  French  W 
Indies,  Netherlands,  Frencl 
Guiana,  aud  Greece,    . 
Argentine  Rep.,  Buenos  Ay 
res,  Chili,   Cuba,  Mexico 
New  Granada,  Peru,  Philip 
pine  Islands,  and  Spain,  . 
Corsica,      France,      Greece 
French     Guiana,    Nether 
lands,   and    French    Wes 
Indies. 

Post  league,    .     .     . 
Judicial  league,    .     . 
Common  league.  . 

4264 
4635 
4861 

2*422 
2'633 
2'761 

H26. 


WEIGHTS  AND  MEASURES. 


19 


Countries. 

Measures. 

Yards. 

English 
mile. 

Australia,    Canada,    Ceylon, 
Corsica,  France,  Gibraltar, 
G.  Britain,  Greece,  British 
Guiana,    French    Guiana, 
N.  Brunswick,  Newfound- 
land, Nova  Scotia,  Nether- 
lands, United  States,  Brit- 
ish W.  Indies,  and  French 
West  Indies,      .... 
Persia,   
Belgium,  Dutch  Guiana,  Hol- 
land, and  Dutch  W.  Indies, 

Marine  or  sea  league, 
Parasang,  .... 

Mile,     

6075 
6086 

6395 

3<451 
3'457 

3<633 

Brazil,  Madeira,  &  Portugal, 

6760 

3'840 

Short  mile, 

6859 

3'897 

Flanders,     

League,      .... 

6864 

3'900 

Argentine  Republic,  Buenos 
Ayres,  Chili,  Cuba,  Mex- 
ico, New  Granada,  Peru, 
Philippine      Islands,     and 
Spain                                  . 

Common  league 

7416 

4'213 

Germany,    

Geographical  mile,  . 

8100 

4'602 

Poland         

Long  mile,      . 

8100 

4'602 

Wertemburg,   

Mile,     

8100 

4'602 

Prussia,  
Denmark,  Hamburg,  Norway, 
and  Danish  West  Indies, 
Vienna 

Post  mile 

8237 

8244 
8296 

4'680 

4<684 
4;713 

Dantzic,       

Mile,     

8475 

4*815 

Hungary,     

9113 

5'  177 

Switzerland,     
Saxony,  

League,      .... 
Mile,     

9153 
9914 

5'200 
5'632 

Germany,    

Lonf  mile, 

10126 

5'753 

10137 

5'759 

Hanover 

Mile            .          .     . 

11559 

6'567 

Swcdc.n,  and  Swedish  West 
Indies,      

League,      .... 

11700 

6'647 

DEFINITIONS. 


^T  27.  A  Line  is  that  which  has 
length,  without  breadth  or  thickness. 
Thus,  AB  and  CD  are  lines. 


A  Superficies  or  Surface  is  a  figure 
that  has  length  and  breadth,  without 
thickness.  Thus  ABCD  is  a  superfi- 
cies or  surface. 


A  Solid  i^  a  figure  that  has  length, 
breadth,  and  thickness.  Thus,  ABC 
DEFGH  is  a  solid. 


A  B 

Magnitude  is  that  which  has  one  or  more  of  the  three  di- 
mensions, length,  breadth,  and  thickness. 

Mensuration  is  the  art  of  measuring  lines,  surfaces,  and 
solids.  It  is  divided  into  three  general  classes,  Lineal,  Super- 
ficial, and  Solid  measures. 

Lineal  Measure  is  the  measure  of  length. 

Superficial  Measure  is  the  measure  of  length  and  breadth, 
or  of  surface. 

Solid  Measure  is  the  measure  of  length,  breadth,  and  thick- 
ness, or  of  solidity. 

IT  27.  Topic.  A  line.  A  superficies  or  surface.  A  solid.  Magnitude. 
Mensuration.  Lineal  measure.  Superficial  measure.  Solid  measure.  Note. 
Classification  of  mensuration.  Geometry.  Geometrical  figures. 


IT  28. 


GEOMETRICAL  DEFINITIONS. 


21 


NOTE.  Superficial  and  solid  measures  are  used  only  in  estimating  the  su- 
perficial and  solid  contents  of  figures,  the  dimensions  of  the  figures  always 
being  taken  in  lineal  measure. 

Mensuration  in  its  various  forms  is  classed  under  that  branch 
of  mathematics  called  Geometry.  Hence, 

Geometry  is  the  science  of  magnitude  in  general.  The  fig- 
ures generally  considered  in  Mensuration  are  called  Geometri~ 
cal  Figures. 


GEOMETRICAL  DEFINITIONS. 

LINES  AND  ANGLES. 

*R"  98.     1.  A  Point  is  that  which  has  position,      ^ 
but  not  magnitude.     Thus,  A  is  a  point. 

2.  A  Line  may  be  either  right  (straight)  or  curved. 

3.  A  Right  Line  is  the  shortest  distance 
that  can  be   drawn  between  two  points.     A 
Thus,  AB  is  a  right  line. 


4.  A  Curve  Line  is  that  which  is  neither 
a  right  line  nor  composed  of  right  lines. 
Thus,  AB  and  CD  are  curve  lines. 


NOTE  1.    A  right  line  is  commonly  called  a  line>  and  a  curve  line  a  curve. 

5.  Parallel  Lines  are  those  which  run 

in  the  same  direction,  at  an  equal  distance     -4 
from  each  other,  and  never  meet.  Thus,  the 
lines  AB  and  CD  are  parallel  to  each  other. 

6.  Parallel   or    Concentric    Curves   are 
those  which  are  equally  distant  from  each 
other  at  every  point.     Thus,  the  curves 
AB  and  CD  are  parallel  to  each  other. 

7.  A  Horizontal  Line  is  a  line  drawn 
parallel   to  the  horizon.     Thus,  the   line 
AB  is  horizontal. 


8.  A  Vertical  Line  is  one  which  ex- 
tends in  a  right  line  from  some  point  to- 
wards the  center  of  the  earth.  Thus,  the 
line  CD  is  vertical. 


D 


IT  28.    Topic.    A  point.    Aline.    A  right  line.    A  curve  line.    Note  1. 
Parallel  lines.    Parallel  or  concentric  curves.    A  horizontal  line.    A  vertical 


22 


GEOMETRICAL  DEFINITIONS. 


IT  28. 


D 


9.  One  line  is  said  to  be  Perpendicular 
to  another,  when  it  so  stands  upon  the 
other  as  to  incline  to  neither  side.     Thus, 
the  line  CD  is  perpendicular  to  the  line 
AB. 

10.  Oblique  Lines  are  those  which  con- 
tinually approach  each  other.     Thus,  the 
lines  AB  and  CD  are  oblique. 

11.  An  Angle  is  the  space  comprised 
between  two  lines  that  meet  in  a  point. 
The  point  of  meeting  is  the  Vertex  of  the 
angle,  and  the  lines  containing  the  angle 
are  its  Sides  or  Legs.     Thus,  the  space 
comprised  between  the  lines  AB  and  CB 
is  an  angle  ;  the  point  B  is  its  vertex ;  and 
the  lines  AB  and  CB  are  its  sides  or  legs. 

NOTE  2.  An  angle  is  generally  read  by  placing  the  letter  at  the  vertex  in 
the  middle  ;  thus,  the  angle  A  EC.  Or,  the  letter  at  the  vertex  only  may  be 
named  ;  thus,  the  angle  B. 

12.  A  Right  Angle  is  one 
formed  by  one  right  line  falling 
on     another     perpendicularly. 
Thus,  ABC  is  a  right  angle. 

13.  An    Obtuse     Angle    is 
greater   than    a    right    angle. 
Thus,  ABD  is  an  obtuse  angle. 

14.  An  Acute  Angle  is  less  than  a  right  angle, 
is  an  acute  angle. 

NOTE  3.     Obtuse  and  acute  angles  are  also  called  Oblique  Angles. 

15.  A  Rectilinear  or  Right-Lined  Angle  is  formed  by  two 
lines.     It  may  be  right,  obtuse,  or  acute.     Thus,  ABC,  ABD, 
and  ABE,  are  rectilinear  angles. 


16.  A  Curvilinear  Angle  is  formed  by 
two  curves.  Thus,  BAG  is  a  curvilinear 
angle. 


line.     Lines  perpendicular  to  each  other.     Oblique  lines.    An  angle.    Its 
vertex.    Its  sides  or  legs.     Note  2.    A  right  angle.    An  obtuse  angle. ^A 
acute  angle.     Note  3.    A  rectilinear  or  right-lined  angle.    A  curvilinear 
angle.    A  mixed  angle.    Adjacent  or  contiguous  angles. 


If  29, 30.  GEOMETRICAL  DEFINITIONS. 


23 


17.  A  Mixed  Angle  is  formed 
by  a  line  and  a  curve.     Thus, 
ABC  is  a  mixed  angle. 

18.  Adjacent  or  Contiguous 
Angles  are  such  as  have  one 
leg  common  to  both  angles. 
Thus,  the  angles  ABD  and 
DBC  are  contiguous. 


PLANE  FIGURES. 

5F  29.     1 .  Plane  Figures  are  even  or  level  surfaces,  bounded 
on  all  sides  by  lines  or  curves. 

2.  Rectilinear  Plane  Figures  are  planes  bounded  by  lines. 

3.  Curvilinear  Plane  Figures  are  planes  bounded  by  curves. 

4.  Mixtilinear  Plane  Figures  are  planes  bounded  by  lines 
and  curves. 

RECTILINEAR  PLANE  FIGURES. 

IF  3O.     1.  Rectilinear  plane  figures  are  called  Polygons. 

2.  A  Regular  Polygon  is  one  whose  sides  are  all  equal. 

3.  An  Irregular  Polygon  is  one  whose  sides  are  unequal. 

4.  The  Perimeter  of  a  polygon  is  the  sum  of  all  its  sides, 
or  the  distance  round  it. 

5.  Similar  Rectilinear  Figures  are  such  as  have  their  sev- 
eral angles  respectively  equal  each  to  each,  and  their  sides 
about  the  equal  angles  proportional. 


6.  A  Triangle  is  a  polygon  of  three 
sides.     Thus,  ABC  is  a  triangle. 


IT  29.  Topic.  Plane  figures.  Rectilinear  plane  figures.  Curvilinear 
plane  figures.  Mixtilinear  plane  figures. 

IT  30.  Topic.  Polygons.  A  regular  polygon.  An  irregular  polygon. 
Perimeter  of  a  polygon.  Similar  rectilinear  figures.  Triangle.  Quadri- 


GEOMETRICAL  DEFINITIONS. 


IT  30. 


7.  A  Quadrilateral  is  a  polygon  of 
four  sides.  Thus,  ABCD  is  a  quadri- 
lateral. 


8.  A  Pentagon  is  a  polygon  of  five       A 
sides.    Thus,  ABODE  is  a  pentagon. 


9.  A  Hexagon  is  a  polygon  of  six 
sides.  Thus,  ABCDEF  is  a  hexa- 
gon. 


10.  A  Heptagon  is  a  polygon  of 
seven  sides.  Thus,  ABCDEFG  is  a 
heptagon. 


G F 

11.  An  Octagon  is  a  polygon  of        X{ 
WEFGH  is  ; 


eight  sides.     Thus,  ABC 
an  octagon 


\_> 

YfrN         ,..      .......       ^y 


lateral.    Pentagon.    Hexagon.    Heptagon.    Octagon.    Nonagon.    Decagon. 


IT  30. 


GEOMETRICAL  DEFINITIONS. 


12.  A  Nonagon  is  a  polygon  of  nine  sides. 

13.  A  Decagon  is  a  polygon  of  ten  sides. 

14.  An  Undecagon  is  a  polygon  of  eleven  sides. 

15.  A  Dodecagon  is  a  polygon  of  twelve  sides. 
Triangles  are  distinguished  as  Right-angled,  Obtuse-angled, 

Acute-angled,  Equilateral,  Isosceles,  and  Scalene. 

16.  A  Right-angled  Triangle  has 
one   right  angle.     Thus,   ABC  is   a 
right-angled  triangle. 

NOTE   1.     A   right-angled  triangle  is   also 
called  a  Rectangular  Triangle. 


17.  An  Obtuse-angled  Triangle  has 
one  obtuse  angle.  Thus,  ABC  is  an 
obtuse-angled  triangle. 


18.  An  Acute-angled  Triangle  has 
all  the  three  angles  acute.  Thus, 
ABC  is  an  acute-angled  triangle. 

NOTE  2.  Obtuse-angled  and  acute-angled 
triangles  are  also  called  Oblique-angled  Trian- 
gles. 


19.  An  Equilateral  Triangle  has 
all  the  three  sides  equal.  Thus,  ABC 
is  an  equilateral  triangle. 


20.  An  Isosceles  Triangle  has  only 
two  of  its  sides  equal.  Thus,  ABC  is 
an  isosceles  triangle. 


Undecagon.  Dodecagon.  Classification  of  triangle 
Note  1.  Obtuse-angled  triangle.  Acute-angled  tri 
eral  triangle.  Isosceles  triangle.  Scalene  triangle. 

3 


ight-angled  triangle. 

Note  2.    Equilat- 

llelogram.    Square. 


26  GEOMETRICAL  DEFINITIONS.  1!  30. 


21.  A  Scalene  Triangle  has  all  the 
three  sides  unequal.  Thus,  ABC  is  a 
scalene  triangle. 


22.  A  Parallelogram  is  a  right-lined  figure,  whose  opposite 
sides  are  parallel,  and  consequently  equal. 

23.  A  Square  is  a  figure  having  four 
equal  sides  and  four  right  angles.     It 
is  a  parallelogram  whose  sides  are  all 
equal,  and  whose  angles  are  all  right 
angles.     Thus,  ABCD  is  a  square. 

24.  A  Rectangle  is  a  right-angled 
parallelogram,  whose  length  exceeds 
its  breadth.     Thus,  ABCD  is  a  rect- 
angle. 

NOTZ    3.      The    areas    of  rectangles    and 
squares  are  sometimes  called  rectangles.  ^  •  ^ 

25.  An  Equilateral  Figure  is  one  that  has  all  its  sides  equal 
to  each  other ;  as,  the  square,  the  equilateral  triangle,  and  all 
the  regular  polygons. 

26.  An  Equiangular  Figure  is  one  that  has  all  its  angles 
equal  to  each  other ;  as  all  the  regular  polygons. 

27.  A  Quadrilateral  Figure  is  one  contained  by  four  right 
lines;  as,  the  square,  the  rectangle,  &c. 

2S.  A  Rhombus  or  Rhomb  is  an 
oblique-angled  equilateral  parallelo- 
gram. It  is  a  quadrilateral  whose 
sides  are  equal,  and  the  opposite  sides 
parallel,  but  the  angles  unequal,  two 
being  obtuse  and  two  acute.  Thus, 
ABCD  is  a  rhombus. 

29.  A  Rhomboid  is  an  oblique-an- 
gled parallelogram.  It  is  a  quadrilat- 
eral whose  opposite  sides  and  angles 
are  equal,  but  which  are  neither  equi- 
lateral nor  equiangular.  Thus,  ABCD 
is  a  rhomboid. 


Rectangle.    Note  3.    An  equilateral  figure.     An  equiangular  figure.      A 


IT  31. 


GEOMETRICAL  DEFINITIONS. 


30.  A  Trapezoid  is  a  quadrilateral 
which  has  two  of  the  opposite  sides 
parallel.  Thus,  ABCD  is  a  trapezoid. 


31.  A  Trapezium  is  a  quadrilateral 
which  has  not  two  sides  parallel. 
Thus,  ABCD  is  a  trapezium. 


32.  A  Diagonal  is  a  line  drawn 
through  a  figure,  joining  two  opposite 
angles.  Thus,  AC  is  the  diagonal  of 
the  rectangle  ABCD. 


33.  The  Apex  of  a  figure   is  its 
highest  point.     Thus,  C  is  the  apex 
of  the  triangle  ABC. 

34.  The  Altitude  of  a  figure  is  the 
perpendicular  hight  of  its  apex  above 
its  base.     Thus,  DC  is  the  altitude  of 
the  triangle  ABC. 


CURVILINEAR  AND  MIXTILINEAR  PLANE  FIGURES. 


ure 


31.     1.  A  Circle  is  a  plane  fig- 
comprehended by  a  single  curve, 


, 

called  its  Circumference  or  Periphery, 
every  part  of  which  is  equally  distant 
from  a  point  called  the  Center.  Thus, 
the  space  inclosed  by  the  curve  ACE 
is  a  circle,  the  curve  is  the  circumfer- 
ence or  periphery,  and  the  point  F  is 
the  center. 


quadrilateral  figure.    Rhombus,  or  rhomb. 
zium.    Diagonal.    Apex.    Altitude. 


Trapezoid.    Trape- 


28 


GEOMETRICAL  DEFINITIONS. 


1T31. 


NOTE  1 .    The  circumference  of  a  circle,  for  the  sake  of  brevity,  is  frequently 
called  a  circle. 

2.  The  Diameter  of  a  circle  is  a  line 
passing  through  the  center,  and  ter- 
minating at  each  end  in  the  circum- 
ference.    It  divides  the  circle  into  two 
equal  parts,  called  Semi-circles.    Thus, 
AD  is   the    diameter    of   the   circle 
ABDE,   and  ABD    and    AED  are 
semi-circles. 

3.  The  Radiics  of  a  circle  is  a  line 
extending  from  the  center  to  the  pe- 
riphery.   It  is  the  semi-diameter.    Two 
or  more  such  lines  are  called  Radii. 
All  radii  of  a  circle  are  equal  to  each 
other.     Thus,  AF,  CF,  DF,  and  EF, 
are  radii  of  the  circle  ACDE.  and  are 
all  equal  to  each  other. 

4.  An  Arc  is  any  part  of  the  cir- 
cumference of  a  circle.     Thus,  GEH 
is  an  arc. 

5.  A  Chord  is  a  line  joining  the  two 
extremities  of  the  arc  of  a  circle.     It 
divides   the  circle   into  two   unequal 
parts.     Thus,  GH  is  a  chord. 

6.  A  Segment  is  that  part  of  the  area  of  a  circle  contained 
between  an  arc  and  its  chord.     It  is  the  part  of  a  circle  cut  off 
by  a  chord.     Thus,  the  space  GHE  is  a  segment. 


7.  A  Sector  is  a  part  of  a  circle 
comprehended  between  two  radii  and 
the  included  arc.  Thus  AFC  and 
CFH  are  sectors. 


IT  31.    Topic.    A  circle.     Bounding  line.     Center.     Note.     Diameter. 
Semi-circle.    Radius.    Radii.     Their  equality.    An  arc.    A  chord.     Seg- 


U  31. 


GEOMETRICAL  DEFINITIONS. 


29 


8.  A  Quadrant  is  the  quarter  of  a 
circle,  or  of  the  circumference  of  a 
circle.  Thus,  AFE  and  EFD  are 
quadrants. 


9.  A  Sextant  is  the  sixth  part  of  a 
circle.  Thus,  AFB,  BFC,  and  CFD, 
are  sextants. 


10.  The    circumference   of  every 
circle  is  divided  into  360  equal  parts, 
called  Degrees ;  each  degree  into  60 
equal  parts,  called  Minutes  ;  and  each 
minute  into  60  equal  parts,  called  Sec- 
onds. 

11.  Degrees,  minutes,  and  seconds, 
are  marked  respectively  °,  ',  " ;  they 
are  used  in  mensuration  and  geometry, 
for  the  measurement  of  angles. 

12.  Every  semi-circle   contains  180°,  every  quadrant  90°, 
and  every  sextant  60°. 

13.  If  two  lines  perpendicular  to  each  other  cross  in  the 
center  of  a  circle,  and  terminate  in  its  circumference,  they  will 
divide  the  circle  into  four  equal  parts,  or  quadrants,  each  hav- 
ing a  right  angle  at  the  center.     Hence,  every  right  angle  con- 
tains 90  degrees. 


14.  Concentric  Circles  are  circles  of 
different  radii,  having  a  common  cen- 
ter. Thus,  abc  and  def  are  concen- 
tric circles. 


270° 


ment.     Sector.     Quadrant.     Sextant.     Divisions  of  the  circumference  of 
circles.    Signs,    Use  of  °,  ,  and  ".    Number  of  degrees  in  a  circle  ;  —  in  a 


30 


GEOMETRICAL  DEFINITIONS. 


15.  An  Ellipse  is  an  oval 
figure,  bounded  by  one  con- 
tinuous curve.  It  has  two 
diameters,  the  longer  of 
which  is  called  the  Trans- 
verse, and  the  shorter  the 
Conjugate  diameter.  The 
two  diameters  are  also  called 
the  Axes.  Thus,  AC  is  the 
transverse,  and  BD  the  con- 
jugate diameter  of  the  ellipse 
ABCD. 

16.  A  circle,  so  drawn  in  a  polygon 
that  its  periphery  touches  all  the  sides 
of  the  polygon,  is  said  to  be  inscribed 
in  the  polygon,  and  the  polygon  is  said 
to  be  circumscribed  about  the  circle. 
Thus,  the  circle  abcdef  is  inscribed 
in  the  polygon  ABCDEF,  and  the 
polygon  is  circumscribed  about  trie 
circle. 

17.  A  polygon,  so  drawn  in  a  circle 
that  each  of  its  angles  stands  on  the 
periphery  of  the  circle,  is  said  to  b 
inscribed  in  the  circle,  and  the  circle 
is  said  to  be  circumscribed  about  the 
polygon.  Thus,  the  polygon  abed  ef 
is  inscribed  in  the  circle,  and  the  circle 
is  circumscribed  about  the  polygon. 

NOTE  2.    Each  of  the  regular  polygons  may  be  inscribed  in,  or  circum- 
scribed about,  a  circle. 

SOLIDS  OR  BODIES. 
IT  32.     1.  A  Solid  or  a  Body  is  a  magnitude  which  has 


by  many  faces  or  plane, 

u  a  quadrant;  -  in  a  sextant.    Proof  that  every  right  angle  co7- 
Concentric  circles.    An  ellipse.    Itsdiame.ers.    Its  axes      C.rcle 
a     o.o,,    Polygon  circuscribed  about  a  "-Polygon 

-. 


ir.32. 


GEOMETRICAL  DEFINITIONS. 


31 


3.  A  Regular  Solid  is  one  whose  faces  are  all  regular  poly- 
gons, similar  and  equal  to  each  other. 

4.  A  Solid  Angle  is  one  made  by  the  meeting  of  more  than 
two  plane  surfaces  at  one  point. 

5    Similar  Solids  are  such  as  are  contained  by  the  same  nur 
ber  of  similar  planes,  similarly  situated,  and  having  h      angles. 


6.  A  Prism  is  a  solid  whose  bases 
or  ends  are  any  similar,  equal,  and 
parallel  plane  figures,  and  -whose  sides 
are  parallelograms. 


7.  A  Cube  is  a  solid  bounded  by  six 
equal  squares.  The  cube  is  some- 
times called  the  Right  Prism. 


S.  A  Parallelepiped  is  a  solid 
bounded  by  six  parallelograms,  the 
opposite  ones  of  which  are  parallel  and 
equal  to  each  other.  Or,  it  is  a  prism 
whose  base  is  a  parallelogram. 

9.  A  Cylinder  is  a  long,  circular 
body,  of  uniform  diameter,  its  extrem- 
ities being  equal  parallel  circles. 

10.  A  Cylindroid  is  a  solid  which 
differs  from  the   cylinder   in  having 
ellipses  instead  of  circles  for  its  ends 
or  bases. 


11.  A  Pyramid  is  a  solid  whose 
base  is  a  polygon,  and  whose  sides  are 
triangles  terminating  in  a  point  called 
the  Vertex. 


32 


GEOMETRICAL  DEFINITIONS. 


12.  The  Segment  of  a  Pyramid  is  a 
part  cut  off  by  a  plane  parallel  to  the 
pyramid's  base. 


13.  TJie  Frustrum  of  a  Pyramid  is 
the  part  left,  after  cutting  off  a  seg- 
ment. 


14.  A  Cone  is  a  solid  whose  base  is 
a  circle,  and  whose  top  terminates  in 
a  point  or  vertex. 


15.  The  Segment  of  a  Cone  is  a  part 
cut  off  by  a  plane  parallel  to  the  cone's 
base. 


16.  The  Frustrum  of  a  Cone  is  the 
part  left,  after  cutting  off  a  segment. 


17.  A  Sphere  or  Globe,  is  a  solid 
bounded  by  a  single  surface,  which  in 
every  part  is  equally  distant  from  a 
point  called  its  center. 


angle.  Similar  solids.  Prism.  Cube.  Parallelepiped.  Cylinder.  Cylin- 
droid.  Pyramid.  Its  vertex.  Segment  of  a  pyramid.  Frustrum  of  a  pyr- 
amid. Cone.  Segment  of  a  cone.  Frustrum  of  a  cone.  Sphere  or  globe. 


f  32.  GEOMETRICAL  DEFINITIONS.  33 


18.  The  Axis  of  a  Sphere  is  a  right 
line,  real  or  imaginary,  passing 
through  its  center,  on  which  it  does 
or  may  revolve. 


19.  The  Diameter  of  a  Sphere  is  a  right  line  passing  through 
its  center,  and  terminating  at  its  surface. 

20.  The  Radius  of  a  Sphere  is  its  semi-diameter. 

If  a  sphere  be  divided  into  two  equal  parts,  by  a  plane  pass- 
ing through  its  center,  the  parts  will  be  called  Hemispheres. 
Hence, 


21.  A  Hemisphere  is  one  half  of  a 
sphere  or  globe. 


The  regular  solids  are  five  in  number ;  the  Tetraedron,  the 
Hexaedron,  the  Octaedron>  the  Dodecaedron,  and  the  Icosaedron. 


22.  The  Tetraedron  is  a  triangular 
pyramid,  bounded  by  four  equal  and 
equilateral' triangles. 


23.  The  Hexaedron  or  Cube  is  a  solid  bounded  by  six  equal 
squares. 

Its  axis.    Its  diameter.    Its  radius.    Hemisphere.    Classification  of  the  reg- 


34 


GEOMETRICAL  DEFINITIONS. 


IF  32. 


24.  The  Octaedron  is  a  solid  bound- 
ed by  eight  equal  and  equilateral  tri- 
angles. 


25.  The  Dodecaedron  is  a  solid 
bounded  by  twelve  equal  regular  pen- 
tagons. 


26.  The  Icosaedron  is  a  solid  bound- 
ed by  twenty  equal  and  equilateral  tri- 
angles. 


NOTE  1.    Each  of  the  regular  solids  may  be  so  contained  within  a  sphere 
that  its  angles  would  all  stand  on  the  superficies  of  the  sphere. 

NOTE  2.    All  the  angles  of  a  regular  solid  must  be  equal  to  each  other. 


ular  solids.    Tetraedron.    Hexaedron.     Octaedron.    Dodecaedron.    Icosae- 
dion.    Notel.    Note  2. 


f  33-35.  PRACTICAL  GEOMETRY.  35 


PRACTICAL  GEOMETRY. 

^T  33.  A  Problem  is  a  proposition  or  a  question  proposed, 
which  requires  some  operation  to  be  performed ;  as,  to  describe 
or  draw  any  of  the  Geometrical  figures. 

Performing  the  operation  is  called  Solving  the  'problem. 

Practical  Geometry  explains  the  methods  of  constructing  or 
describing  the  geometrical  figures. 

IT  34.  Some  instruments  will  be  necessary  to  the  suc- 
cessful prosecution  of  this  subject.  A  case  of  drafting  instru- 
ments will  best  answer  the  purpose,  but  when  these  cannot  be 
obtained,  the  dividers  or  compasses,  a  common  ruler,  and  a 
scale  of  equal  parts,  will  be  found  sufficient  for  the  solution  of 
all  the  geometrical  problems  contained  in  this  work. 

The  dividers  are  so  well  known  that  a  description  of  them  is 
deemed  unnecessary. 

The  ruler  may  be  any  convenient  length  from  12  to  18 
inches,  from  1  to  2  inches  in  width,  and  from  J  to  §  of  an  inch 
in  thickness. 

The  scale  of  equal  parts  may  be  conveniently  constructed  on 
one  side  of  the  common  ruler,  as  follows  :  Lay  off  any  portion 
of  one  side  of  the  ruler,  say  10  inches,  into  10  equal  parts,  thus 
making  each  part  T\y  of  the  length  of  the  scale,  or  1  inch  in 
length.  Number  these  parts  in  their  order  from  left  to  right; 
thus,  1,  2,  3,  4,  &c.  Then  lay  off  one  of  these  parts  into  10 
other  equal  parts,  each  part  being  -fa  of  an  inch,  or  ^-^  of  the 
length  of  the  scale.  Number  these  parts  in  their  order  from 
left  to  right,  and  the  scale  will  be  completed. 


12  3456789 


|  _ 

1  1  1  1  1  1  M  1  1  —  '  -  1  -  — 


4 


If  35.    Geometrical  Problems. 

PROBLEM  I. 
To  draw  a  line  through  a  given  point  parallel  to  a  given  line.  ' 

If  33.    Topic.    A  problem.     Solving  a  problem.     Practical  Geometry. 

IT  34,  Topic.  Instruments  necessary  for  the  solution  of  the  geometrical 
problems.  The  dividers  or  compasses.  The  ruler.  Construction  of  the 
scale  of  equal  parts. 


36 


GEOMETRICAL  PROBLEMS. 


Let  AB  be  the  given  line,  and  C  the         # 

given  point.    With  C  as  a  center,  and  any     ^~T~~" 

convenient  radius  greater  than  the  shortest  \ 

distance  from  C  to  AB,  as  CD,  describe  an        — \  , 

arc  DF  indefinitely.     With  the  same  radius, 

and  D  as  a  center,  describe  the  arc   CG.     Then  make  DF=  CG,  and 

draw  the  line  CF,  which  will  be  parallel  to  AB. 

PROBLEM  II. 
To  bisect  a  given  line,  or  to  divide  it  into  two  equal  parts. 


Let  AB  be  the  given  line.  With  A  as  a 
center,  and  any  radius  greater  than  half  of 
AB,  describe  arcs  above  and  below  AB,  as 
at  C  and  D.  With  the  same  radius,  and  B 
as  a  center,  describe  arcs  above  and  below 
AB,  intersecting  the  arcs  first  drawn,  at  C 
and  D.  Draw  the  line  FG  through  the 
points  C  and  D,  and  it  will  divide  the  line 
AB  at  E)  into  two  equal  parts  AE  and  BE. 


PROBLEM  III. 
To  bisect  a  given  curve. 


Let  AB  be  the  given  curve.  With  A 
and  B  as  centers,  and  any  radius  greater 
than  half  of  AB,  describe  arcs  above  and 
below  AB,  intersecting  each  other  at  C  and 
D.  Draw  the  line  FG  through  the  points 
C  and  D,  and  it  will  bisect  the  curve  AB, 
at  E, 


PROBLEM  IV. 
To  bisect  a  given  angle. 


Let  BA  C  be  the  given  angle.  Lay  off  upon 
AB  and  AC  two  points,  equally  distant  from  A, 
as  D  and  E.  With  D  and  E  as  centers,  and 
any  radius  greater  than  half  of  DE,  describe 
two  arcs  intersecting  at  jP.  Then  draw  the 
line  AG  through  the  points  A  and  F}  and  it 
will  bisect  the  angle  BAG. 


IF  35.  GEOMETRICAL  PROBLEMS. 

PROBLEM  V. 
To  erect  a  perpendicular  on  the  middle  of  a  given  line. 


Let  AB  be  the  given  line.  Bisect  the 
line  AB,  by  Prob.  II.  Then  the  line  FE 
will  be  perpendicular  to,  and  will  stand  on 
the  middle  of  the  line  AB, 


PROBLEM  VI. 
To  erect  a  perpendicular  on  any  given  point  in  a  line. 

Let  E  be  the  given  point,  and  AB  the 
given  line.     From  E  lay  off  any  two  equal 
distances,  EG  and  EH,  upon  the  line  AB. 
With  G  and  H  as  centers,  and  any  radius 
greater  than  EG,  describe  two  arcs  inter- 
secting each  other  in  C.     Then  draw  the      * 
line  FE,  and  it  will  be  the  required  perpen-        ~~& 
dicular. 

SECOND  METHOD. 


Let  B  be  the  given  point,  and  AB  the 
given  line.  With  any  point  C  as  a  center, 
and  a  radius  equal  to  BC,  describe  the  semi- 
circle DBE.  Draw  the  diameter  DE 
through  the  points  D  and  C.  Then  draw  a 
line  from  B  through  the  point  E,  and  it  will 
be  the  required  perpendicular. 


NOTE.  The  second  method  of  solving  this  problem  is  based  upon  the  prin- 
ciple that  all  angles  in  a  semicircle  are  "right  angles.*  In  erecting  a  perpen- 
dicular on  or  near  the  end  of  a  line,  the  second  method  is  preferable  to  the 
first. 

PROBLEM  VII. 

From  any  point  without  a  given  line  to  draw  a  perpendicular  to 
the  line. 


*  Euclid's  Elements  of  Geometry. 


GEOMETRICAL  PROBLEMS. 


1135. 


Let  A  be  the  given  point,  and  EC  the 
given  line.  With  A  as  a  center,  and  any 
radius  greater  than  the  shortest  distance 
from  A  to  the  line  EC,  describe  an  arc  in- 
tersecting EC  in  two  points,  D  and  E, 
which  are  equi-distant  from  A.  With  D  ^ 
and  E  as  centers,  and  the  radius  AD,  de- 
scribe two  arcs  intersecting  each  other  in  F. 
Then  draw  the  line  AF,  and  AG  will  be 
the  required  perpendicular. 

PROBLEM  VIII. 

To  describe  a  circle  which  shall  pass  through  any  three  given 
points  not  in  a  right  line. 


Let  A,  B,  and  C  be  the  given  points.  Connect 
the  points  A  and  B,  and  the  points  B  and  C,  by  the 
lines  AB  and  EC.  Bisect  the  lines  AB  and  EC, 
by  Prob.  II.,  and  the  point  D,  where  the  bisecting 
lines  cross  each  other,  will  be  the  center  of  the  cir- 
cle. Then  with  the  radius  DA,  DB,  or  DC,  de- 
scribe a  circle  which  will  pass  through  the  points 
A,  B,  C. 

PROBLEM   IX. 
To  find  the  center  of  a  circle. 


Take  any  three  points  in  the  circumference,  as 
A,  B,  C,  and  connect  them  by  the  chords  AB  and 
EC.  Bisect  the  chords  AB  and  EC  by  Prob.  II., 
and  the  point  D,  where  the  bisecting  lines  cross 
each  other,  will  be  the  center  of  the  circle. 


PROBLEM   X. 

To  find  the  center  of  a  circle  of  which  an  arc  only  is  given. 

Let  AC  be  the  given  arc.  Take  any  point 
in  the  arc,  as  B,  and  connect  it  with  the  ex- 
tremities of  the  arc  by  the  chords  AB  and  EC. 
Bisect  these  chords  by  Prob.  II.,  and  the  point 
D,  where  the  bisecting  lines  cross  each  other, 
will  be  the  center  of  the  circle. 


IT  35. 


GEOMETRICAL  PROBLEMS. 


39 


PROBLEM  XI. 

To  draw  a  curve  through  a  given  point  parallel  to  a  given 

curve. 

Let  AB  be  the  given  curve  or  arc, 
and  C  the  given  point.  First  find  the 
center  of  the  circle  of  which  the  curve 
AB  is  an  arc,  by  Prob.  X.  Then,  with 
D  as  a  center,  and  a  radius  equal  to  DC, 
describe  the  arc  EF,  which  will  be  par- 
allel to  the  arc  AB. 

PROBLEM  XII. 

The  base  and  perpendicular  of  a  right-angled  triangle  being 
given,  to  describe,  the  triangle. 


Let  D  be  the  given  base,  and  E  the  per- 
pendicular. Draw  the  base  AB  equal  to 
the  line  D.  Upon  the  point  B  erect  the 
perpendicular  BC,  equal  to  the  line  E,  by 
Prob.  VI.,  and  draw  the  line  AC.  Then 
the  triangle  ABC  will  be  the  required  tri- 
angle. 


PROBLEM  XIII. 
To  describe  an  equilateral  triangle  upon  a  given  line  or  side. 


Let  AB  be  the  given  line  or  side.  With  A 
and  B  as  centers,  and  the  radius  AB,  describe 
two  arcs  intersecting  each  other  in  C.  Then 
draw  the  lines  AC  and  BC  and  ABC  will  be 
the  required  triangle. 


PROBLEM  XIV. 
The  three  sides  of  a  triangle  being  given,  to  describe  the  triangle. 


Let  A,  B,  and  C  be  the  given  sides. 
Draw  DE  equal  to  the  line  A.  With  D  as 
a  center,  and  a  radius  equal  to  the  line  B, 
and  with  E  as  a  center,  and  a  radius  equal 
to  the  line  C,  describe  arcs  intersecting  each 
other  in  F.  Draw  the  lines  D F  and  EF, 
and  D  EF  will  be  the  required  triangle. 


40 


GEOMETRICAL  PROBLEMS; 


PROBLEM  XV. 

The  hypotenuse  and  one  side  of  a  right-angled  triangle  being 
given,  to  describe  the  triangle. 


Let  D  be  the  given  side,  and  E  the  given 
hypotenuse.  Draw  the  side  AB  equal  to 
the  line  D,  and  upon  the  point  B  erect  the 
perpendicular  B C  indefinitely.  With  A  as  a 
center,  and  a  radius  equal  to  the  line  .E?  de- 
scribe an  arc  intersecting  the  perpendicular 
BC,  at  C.  Then  draw  the  hypotenuse  AC, 
and  ABC  will  be  the  required  triangle. 


D 


PROBLEM  XVI. 
At  a  given  point  to  make  an  angle  equal  to  a  given  angle. 

Let  D  be  the  given  point, 
and  BAG  the  given  angle. 
Draw  the  line  DE  indefinitely. 
With  A  as  a  center,  and  any 
convenient  radius,  draw  the 
arc  BC  terminating  in  the 
sides  of  the  angle.  With  the 
same  radius,  and  D  as  a  center, 
draw  the  arc  EF.  With  E  as 

a  center,  and  a  radius  equal  to  BC,  draw  an  arc  intersecting  the  arc  EF 
at  F.  Then  through  the  points  D  and  F  draw  the  line  DF}  and  the 
angle  EDF  will  be  equal  to  the  angle  BAG. 

PROBLEM  XVII. 

Two  sides  of  a  triangle  and  the  angle  which  they  contain  being 
given,  to  describe  the  triangle. 

Let  A  and  B  be  the  given  sides,  and 
C  the  given  angle.  Draw  the  side  DE 
equal  to  the  line  A.  At  the  point  E 
make  an  angle  equal  to  the  angle  C,  by 
Prob.  XVI.,  and  draw  the  line  EF  indef- 
initely. With  E  as  a  center,  and  a  ra- 
dius equal  to  the  line  B,  describe  an  arc 
intersecting  the  line  EF  in  G.  Then 
draw  the  line  DG,  and  DEG  will  be  the 
required  triangle. 

PROBLEM  XVIII. 
To  describe  a  square  upon  a  given  line. 


U35. 


GEOMETRICAL  PROBLEMS. 


41 


Let  AB  be  the  given  line.  At  the  point  B  erect 
the  perpendicular  JBC,  and  make  it  equal  to  AB. 
With  A  and  C  as  centers,  and  a  radius  equal  to  AB, 
describe  two  arcs  intersecting  each  other  in  D. 
Then  draw  the  lines  AD  and  CD,  and  ABCD  will 
be  the  required  square. 

PROBLEM  XIX. 

Two  adjacent  sides  of  a  rectangle  being  given,  to  describe  the 
rectangle. 


B 


Let  A  and  B  be  the  given  adjacent  sides. 
Draw  the  side  CD  equal  to  the  line  B,  and 
upon  the  point  D  erect  the  perpendicular  DE 
equal  to  the  line  A.  With  C  as  a  center,  and 
a  radius  equal  to  DE,  describe  an  arc ;  and 
with  E  as  a  center,  and  a  radius  equal  to  CD, 
describe  another  arc,  intersecting  the  first  at 

F.     Then  draw  the  lines   CF  and  EF,  and    A 

CDEF  will  be  the  required  rectangle.  g 

PROBLEM  XX. 

One  side  and  one  of  the  angles  of  a  rhombus  being  given,  to  de- 
scribe the  rhombus. 

Let  AB  be  the  given  side,  and  E  the 
given  angle.  At  the  point  B  make  an 
angle  equal  to  the  angle  E,  by  Prob. 
XVI.,  and  draw  the  line  BC  equal  to  AB. 
With  A  and  C  as  centers,  and  a  radius 
equal  to  AB,  describe  two  arcs  intersect- 
ing each  other  in  D.  Then  draw  the  lines 
AD  and  CD,  and  ABCD  will  be  the  re- 
quired rhombus. 

NOTE.    A  rhomboid  may  be  readily  described,  by  combining  Problems 
XIX.  and  XX. 

PROBLEM  XXI. 
To  inscribe  an  equilateral  triangle  in  a  given  circle. 

With  any  point  in  the  circumference,  as  E, 
for  a  center,  and  the  radius  DE,  describe  two 
arcs  intersecting  the  circle  in  A  and  B .  With  A 
and  B  as  centers,  and  a  radius  equal  to  AB, 
describe  two  arcs  intersecting  each  other  in  C. 
These  arcs  will  intersect  each  other  and  the 
circle  in  the  same  point.  Then  draw  the  lines 
AB,  BC,  and  CA,  and  ABC  will  be  the  re- 
quired triangle. 

4* 


42  GEOMETRICAL  PROBLEMS. 

PROBLEM  XXII. 
To  inscribe  a  square  in  a  given  circle, 


Draw  the  diameters  AC  and  BD  at  right 
angles  to  each  other.  Then  through  the  points 
A,  B,  C,  and  D,  draw  the  lines  AB,  BC,  CD, 
and  DA,  and  ABCD  will  be  the  required 
square. 


PROBLEM  XXIII. 
To  inscribe  a  pentagon  in  a  given  circle. 


Draw  the  diameters  FG  and  EH,  at  right 
angles  to  each  other,  and  bisect  the  radius 
76?  at  K.  With  K  as  a  center,  and  a  radius 
equal  to  EK,  describe  an  arc  intersecting 
FG  in  L.  Apply  the  distance  EL  around 
the  circle,  and  it  will  divide  it  into  five 
equal  parts.  Then  draw  the  lines  AB,  BC, 
CD,  DE,  and  EA,  and  ABCDE  will  be  the 
required  pentagon. 


PROBLEM  XXIV. 
To  inscribe  a  hexagon  in  a  given  circle, 


Apply  the  radius  AG  around  the  circle, 
and  it  will  divide  it  into  six  equal  parts. 
Then  draw  the  lines  A  B,  BC,  CD,  DE,  EF, 
and  FA,  and  ABCDEFwill'be,  the  required 
hexagon. 


K35. 


B 


PROBLEM  XXV. 
To  inscribe  an  octagon  in  a  given  circle. 


Inscribe  the  square  AC  EG,  by  Prob.  XXII., 
and  bisect  the  arcs  AC,  CE,  EG,  and  GA,  at  B, 
D,  F,  and  H,  respectively.  Then  draw  the 
lines  AB,  BC,  CD,  DE,  EF,  FG,  GH,  and 
HA,  and  AB  CD EFGH  will  be  the  required 
octagon. 


GEOMETRICAL  PROBLEMS. 

PROBLEM  XXVI. 
To  inscribe  a  decagon  in  a  given  circle. 


Inscribe  the  pentagon  ABODE,  by  Prob. 
XXIII.,  and  bisect  the  arcs  AB,  BC,  CD, 
DE,  and  EA.  Then  draw  lines  through 
the  angles  of  the  pentagon  and  the  points 
of  bisection,  and  the  figure  will  be  the  re- 
quired decagon. 


PROBLEM  XXVII. 
To  inscribe  a  dodecagon  in  a  given  circle. 

Inscribe  the  hexagon  ABCDEF,  by  Prob. 
XXIV.,  and  bisect  the  arcs  AB,  BC,  CD, 
DE,  EF,  and  FA.  Then  draw  lines 
through  the  angles  of  the  hexagon  and  the 
bisecting  points,  and  the  figure  will  be  the 
required  dodecagon. 

PROBLEM  XXVIII. 

To  inscribe  any  regular  polygon  in  a  given  circle. 
Divide  the  circle  into  as  many  equal  parts  as  the  required  polygon  is 
to  contain  sides,  and  draw  lines  through  the  points  of  division.     The 
inscribed  figure  will  be  the  required  polygon. 

PROBLEM  XXIX. 

One  side  and  the  number  of  sides  of  a  regular  polygon  being 
given,  to  describe  the  polygon. 

Let  it  be  required  to  describe  a  hepta- 
gon upon  the  line  AB.  With  the  center 
A,  and  the  radius  AB,  describe  the  semi- 
circle HabcdefB,  and  divide  it  into  seven 
equal  parts.  To  the  second  point  of  di- 
vision b,  draw  the  line  AG,  and  through 
the  points  c,  d,  e,  and  /,  draw  the  lines 
AF,  AE,  AD,  and  AC.  Apply  the  dis- 
tance AB,  from  B  to  C,  from  C  to  D, 
from  D  to  E,  from  E  to  F,  and  from  F 
to  G.  Then  draw  the  lines  B  C,  CD,  DE, 
EF,  and  FG,  and  ABCDEFG  will  be 
the  required  heptagon. 
Proceed  in  the  same  manner  with  any  other  regular  polygon. 


44 


GEOMETRICAL  PROBLEMS. 


1T35. 


PROBLEM  XXX. 
To  circumscribe  a  regular  -polygon  about  a  given  circle. 

Let  it  be  required  to  circumscribe  a  hexagon 
about  a  circle.  In  the  given  circle  inscribe  the 
hexagon  ABC  DBF.  To  the  length  of  the 
radius  OA,  add  the  distance  Pp,  and  with  this 
radius,  arid  O  as  a  center,  describe  a  second 
circle.  Then  in  this  circle  describe  the  hexagon 
abcdef,  and  it  will  circumscribe  the  given  cir- 
cle. 

Any  other  regular  polygon  may  be  circum- 
scribed about  a  circle  in  the  same  manner. 

PROBLEM  XXXI. 
To  circwnscribe  a  circle  about  a  regular  polygon. 


Bisect  any  two  adjacent  sides  of  the  polygon, 
as  AB  and  BC,  and  the  point  D,  where  the 
bisecting  lines  cross  each  other,  will  be  the 
center  of  the  circle.  Then,  with  D  as  a  center, 
and  a  radius  equal  to  the  distance  from  D  to 
any  angle  of  the  polygon,  as  A,  describe  a  cir- 
cle, and  it  will  circumscribe  the  given  polygon. 


PROBLEM    XXXII. 
To  inscribe  a  circle  in  a  regular  polygon. 


Bisect  any  two  adjacent  sides  of  the  poly- 
gon,  as  AB  and  BC,  and  the  point  D,  where 
the  bisecting  lines  cross  each  other,  will  be 
the  center  of  the  circle.  Then,  with  D  as  a 
center,  and  the  radius  DE,  describe  a  circle, 
and  it  will  be  inscribed  in  the  given  polygon. 


PROBLEM  XXXIII. 

To  inscribe  a  circle  in  a  given  triangle. 
Let  AB  C  be  the  given  triangle. 
Bisect  any  two  angles,  as  A  and  B, 
and  the  point  D,  where  the  bisecting 
lines  cross  each  other,  will  be  the  cen- 
ter of  the  circle.  From  this  point  let 
fall  a  perpendicular  upon  one  of  the 
sides,  as  DE.  Then,  with  the  center 
D,  and  the  radius  DE,  describe  a  cir- 
cle, and  it  will  be  inscribed  in  the 
given  triangle, 


C 


If  35. 


GEOMETRICAL  PROBLEMS. 


45 


PROBLEM  XXXIV. 

To  construct  solids. 

Upon  pasteboard,  or  any  other  pliable  matter,  draw  figures  like  the  fol- 
lowing. Cut  the  bounding  lines  entirely  through,  and  the  other  lines 
half  through ;  turn  up  the  sides  and  glue  the  edges  together,  and  the 
figures  will  form  the  solids  named  below. 


Tetraedron.  Hexaedron. 


Octaedron. 


Dodecaedron. 


Icosaedron.  Parallelopiped^  or  Square  Prism, 


GEOMETRICAL  PROBLEMS. 


1T35. 


Hexagonal  Pyramid.         Frustrum  of  a  Square  Pyramid. 


Cone.  Frustrum  of  a  Com. 


MENSURATION  OF  LINES  AND  SUPERFICIES. 

IT  36      The  area  of  a  figure  is  its  superficial  contents,  or 
the  surface  included  within  any  given  lines,  without  regar 

th  Tn  Taking  the  dimensions  of  any  line,  surface,  or  solid  we 
are  always  governed  by  some  denomination,  a  unit  of  which  is 
called  the  Unit  of  Measure.  Thus,  if  any  lineal  measure  be 
estimated  in  feet,  the  unit  of  measure  is  1  foot;  il:  m  inches, 
the  unit  is  1  inch;  if  in  yards,  the  unit  is  1  yard,  &c. 
any  superficial  measure  be  estimated  m  feet,  the  unit  of  meas- 
ure is  1  square  foot,  or  144  square  inches;  if  m  yards  the 
unit  is  1  square  yard,  or  9  square  feet,  &c.  If  any  solid  or 
cubic  measure  be  estimated  in  feet,  the  unit  of  measure  is  1 
cubic  foot,  or  1728  cubic  inches ;  if  in  yards,  the  unit  is  1  CUDK 
yard,  or  27  cubic  feet,  &c. 

IT  37.     The  length  and  breadth  of  a  square  or  rectangle 
being  given,  to  find  the  square  contents. 

RULE. 

Multiply  the  length  by  the  breadth,  and  the  product  will  be 
the  square  contents. 

NOTE      For  an  analysis  of  the  principles  of  this  and  the  following  rule,  see 
Revised  Arithmetic,  tflT  48,  49,  and  50. 

EXAMPLES  FOR  PRACTICE. 

1.  How  many  square  inches  in  a  board  16  inches  square  ? 

2.  How  many  square  rods  in  a  field  90  rods  long,  andJ2  ™ds^vuk  ? 

H°3W  A  ceryta^nrevmage  lot  of  land  is  66  feet  front,  by  330  feet  deepj  how 
many  poles  does  it  contain  ? 

4.  In  a  field  220  rods  long  and  90  rods  wide,  how  many  acres? 

5    A  certain  rectangular  piece  of  land  measures  1000  links  by  1 
how  many  chains  does  it  contain?    How  many  acres  ? 

IT  38.     The  square  contents  or  area,  and  one  side  of  a 
square  or  rectangle  being  given,  to  fiiul  the  other  side. 

IT  36.    Topic.    The  area  of  a  figure.    Unit  of  measure.    Examples. 
IT  37.    Topic.    Analysis.    Rule. 


48  MENSURATION  OF  LINES  IT  39. 

RULE. 

Divide  the  square  contents  by  the  given  side,  and  the  quo- 
tient will  be  the  required  side. 

NOTE.  The  area  and  the  given  side  must  be  reduced  to  corresponding  de- 
nominations before  dividing ;  that  is,  if  the  area  is  expressed  in  square  feet, 
the  given  side  must  be  in  feet.  Or,  they  may  be  reduced  to  any  other  cor- 
responding denominations,  as  inches  and  square  inches,  yards  and  square  yards, 
rods  and  square  rods,  &c. 

EXAMPLES  FOR  PRACTICE. 

1.  If  a  piece  of  land  20  rods  in  length  contain  240  square  rods,  what 
is  its  width  ? 

2.  The  side  of  a  certain  building  16  feet  in  hight  contains  2560  square 
feet ;  what  is  its  length  ? 

3.  What  length  of.  carpeting  5  quarters  wide   is  equal  to  a  square 
yard?  Ans.  3<2  qrs.  =28'8  in. 

4.  How  many  yards  of  cloth  If  yards  wide  are  equal  to  15  yards  |  of 
a  yard  wide?  Ans,  8-ftyds. 

5.  A  piece  of  land  8  chains  wide  contains  40  acres ;  what  is  its  length 
in  chains  ?  Ans.  50  chains. 

IT  39.  The  base  and  perpendicular  of  a  right-angled  tri- 
angle being  given,  to  find  the  hypotenuse. 

RULE. 

Square  the  base  and  the  perpendicular,  add  the  squares  to- 
gether, and  extract  the  square  root  of  their  sum  ;  the  root  will 
be  the  length  of  the  hypotenuse. 

NOTE  l.  For  an  analysis  of  the  principles  upon  which  this  and  the  follow- 
ing rule  are  founded,  see  Revised  Arithmetic,  it  210,  Note  3. 

EXAMPLES  FOR  PRACTICE. 

1.  The  base  is  12  inches,  and  the  perpendicular  5  inches ;  what  is  the 
hypotenuse  ? 

2.  The  gable  of  a  house  is  28  feet  wide,  and  the  perpendicular  hight 
of  the  ridge  of  the  roof  above  the  eaves  is  7  feet ;  what  is  the  length  of 
the  rafters  ? 

NOTE  2.     The  gable  is  the  portion  above  a  hori- 
zontal  line  extending  from  one  eave  to  the  other. 
Thus,  ABC  may  represent  the  gable  of  a  house, 
and  may  readily  be  divided  into  two  right-angled          _ 
triangles,  ADJB  and  CDB.  Ans.  15'65+  feet.     -«  j)  ** 

3.  Upon  a  plane  25  feet  long  stands  a  pole  12  feet  high,  at  the  distance 
of  9  feet  from  one  end  of  the  plane  ;  what  is  the  length  of  a  rope  that  will 
extend  from  one  end  of  the  plane  to  the  other,  over  the  top  of  the  pole  ? 

Ans.  35  feet. 

IT  38.    Topic.    Analysis.    Rule.     Note. 

TT39.    Topic.    Analysis.    Rule.    Note  1.    Note  2.    Note  3.    Note  4. 


IF  40.  AND  SUPERFICIES.  49 

4.  The  second  floor  of  a  certain  house  is  9  feet  above  the  first,  and 
each  of  the  steps  in  the  flight  of  stairs  leading  from  the  first  floor  to  the 
second,  is  9  inches  high  and  12  inches  wide ;  what  is  the  slant  night  of 
the  flight  of  stairs  ? 

NOTE  3.  The  pupil  will  perceive  that  the  stairs  rise  3-4  as  fast  as  they 
advance  ;  that  is,  the  perpendicular  is  3-4  as  long  as  the  base. 

NOTE  4.  The  hypotenuse  of  any  right-angled  triangle  whose  base  and  per- 
pendicular are  to  each  other  as  4  to  3,  is  equal  to  the  longer  side  plus  i  of  itself. 

5.  The  base  is  20,  and  the  perpendicular  15 5  what  is  the  hypotenuse? 

6.  The  base  is  48,  and  the  perpendicular  64  ;  what  is  the  hypotenuse  ? 

7.  The  hights  of  two  trees,  75  feet  apart,  are  96  and  130  feet ;  how  far 
from  the  top  of  one  tree  to  the  top  of  the  other  ?    How  far  from  the  top  of 
each  to  the  bottom  of  the  other  ? 

!From  top  of  one  to  top  of  the  other,  82<34-}-  ft. 
"  "  "  taller  to  bottom  of  shorter,  150'08  ft. 
"  «  "  shorter  to  bottom  of  taller,  121<82-4-  ft. 

^T  4O.  The  hypotenuse  and  one  leg  of  a  right-angled  tri- 
angle leing  given,  to  find  the  other  leg. 

RULE. 

Square  the  hypotenuse  and  the  given  leg,  subtract  the  square 
of  the  leg  from  the  square  of  the  hypotenuse,  and  extract  the 
square  root  of  the  remainder;  the  root  will  be  the  length  of  the 
other  leg. 

NOTE.  The  pupil  will  perceive  that  this  rule  is  the  reverse  of  the  one  given 
in  the  last  IT. 

EXAMPLES  FOR  PRACTICE. 

1.  The  hypotenuse  of  a  right-angled  triangle  is  5  feet  long,  and  the 
base  4  feet;  what  is  the  length  of  the  perpendicular? 

2.  A  ladder  17  feet  long  is  so  placed  that  it  touches  the  wall  15  feet 
above  the  plane  on  which  the  wall  and  ladder  stand  ;  how  far  from  the 
foot  of  the  wall  to  the  foot  of  the  ladder  ?  Ans.  8  feet. 

3.  The  length  of  the  rafters  to  a  certain  building  is  13  feet,  and  the 
perpendicular  hight  of  the  ridge  above  the  eaves  is  5  feet ;  what  is  the 
width  of  the  gable  ?  Ans.  24  feet. 

4.  One  side  of  the  roof  of  a  certain  house  is   18  feet  wide,  the  other 
side  16,  and  the  perpendicular  hight  of  the  ridge  above  the  eaves  9  feet ; 
what  is  the  width  of  the  gal  '  Ans.  28'8-f-feet. 

5.  A  ladder  25  feet  long  is  so  placed  between  two  buildings,  (hat  when 
its  top  is  leaned  against  one  of  them,  it  touches  the  buikling"20  feet  from 
the  ground,  and  when  leaned  against  the  other,  it  touches  it  15  feet  from 
the  ground ;  what  is  the  horizontal  distance  between  the  buildings  ? 

Ans.  35  feet. 

6.  The  distance  from  the  spot  on  which  I  stand  to  the  top  of  a  certain 
tree  is  100  feet,  and  to  the  bottom  of  the  same,  (which  is  in  the  same 
plane  with  my  feet,)  60  feet ;  how  high  is  the  tree  ?  Ans.  80  feet. 

IT  40.    Topic.    Analysis.    Rule.    Note. 
O 


50  MENSURATION  OF  LINES  ^  41, 42. 

7.  The  distance  from  the  top  of  one  tree  to  the  top  of  "^  "^ 
75  feet  from  the  first,  is  100  feet  and  the  shorter  tree*  80  ieet .  h igh 
what  is  the  hight  of  the  taUer  tree  ? 

f  41 .     TAe  mm  awd  difference  of  two  numbers  being  given, 
to  find  the  numbers. 

NOTE.    A  knowledge  of  the  principles  contained  in  this  and  the  following 
IT  is  necessary  to  a  clear  comprehension  of  the  rule  m  II  43. 

Ex.  The  sum  of  two  numbers  is  25,  and  their  difference  is  7 ; 
what  are  the  numbers  ? 

The  sum  of  two  numbers  plus  their  difference  will  give 


Hence, 

The  sum  and  difference  of  two  numbers  being  given,  to  find 
the  numbers. 

RULE. 

I.  For  the  greater  number  ;  —  Add  the  sum  and  difference 
too-ether,  and  divide  the  amount  by  2. 

II.  For  the  less  number ;  -  Subtract  the  difference  from  the 
sum,  and  divide  the  remainder  by  2. 

EXAMPLES  FOR  PRACTICE. 

1.  The  sum  of  two  numbers  is  92,  and  their  difference  is  56  j  what  are 

th2aThe^um  of  the  lengths  of  two  lines  is  126  yards,  and  their  differ- 
ence is  31  yards  ;  what  is  the  length  of  each  line  ?  ^^  ^  ^ 

Ans-    \  Shorter  «      47<5     « 

3    Two  men  own  350  acres  of  land,  and  one  owns  62  acres  more  than 
the  other ;  how  many  acres  does  each  man  °wnT  ^^  ^^  ^  acres. 

Ans.     j  Tlng  ™tlaier  owns  144      « 

IT  42.  TAe  mm  of  two  numbers  and  the  difference  of  their 
squares  being  given,  to  find  the  numbers. 

Ex  1  The  sum  of  the  numbers  16  and  9  is  25,  and  their 
difference  is  7 ;  what  is  the  difference  of  their  square  > 

ANALYSIS  162  _  256,  9*  =  81,  and  256  -  81  =  175.  But  25,  the  sum 
of  tl^  two  given  Cumbers,  multiplied  by  7,  their  difference,  gives  the 
same'Vesult;  thus,  25  X  7  =  175.  Therefore, _____ 

IT  41.    Topic.     Note.     Solution  of  Ex.  1.     Rule. 

IT  42.    Topic.    Solution  of  Ex.1.    Conclusion.    Solution  of  Ex.  2.    K<  ". 


IT  43.  AND  SUPERFICIES.  51 

The  product  of  the  sum  and  difference  of  two  numbers  is  equal 
to  the  difference  of  their  squares. 

Ex.  2.  The  sum  of  two  numbers  is  25,  and  the  difference 
of  their  squares  is  175 ;  what  are  the  numbers  ? 

ANALYSIS.  In  this  example  we  have  the  sum  of  two  numbers,  and  the 
difference  of  their  squares  given,  to  find  the  numbers.  That  is,  we  have 
the  product,  175,  and  one  factor,  25,  to  find  the  other  factor.  175  -j-  25 
=  7,  the  other  factor,  or  the  difference  of  the  two  numbers.  We  now 
have  25,  the  sum  of  two  numbers,  and  7,  their  difference,  to  find  the 
numbers,  (U"  41.)  25-}-7  =  32,  and  32  -f-  2  =  16,  the  greater  number. 
25  —  7  =  18,  and  18  -5-  2  =  9,  the  less  number.  Hence, 

When  the  sum  of  two  numbers,  and  the  difference  of  their 
squares  are  given,  tojlnd  the  ?mmbers. 

RULE. 

I.  Divide  the  difference  of  the  squares  by  the  sum  of  the 
numbers,  and  the  quotient  '**ill  be  the  difference  of  the  num- 
bers. 

II.  From  the  sum  and  difference,  find  the   numbers  by 
IT  41,  Rule. 

EXAMPLES  FOR  PRACTICE. 

1.  The  sum  of  two  numbers  is  30,  and  the  difference  of  their  squares 
is  300 ;  what  are  the  numbers  ? 

2.  The  sum  of  two  numbers  is  50,  and  the  difference  of  their  squares 
is  112 ;  what  are  the  numbers  ? 

3.  The  sum  of  two  numbers  is  88,  and  the  difference  of  their  squares 
is  6688 ;  what  are  the  numbers  ?  Ans.  82  and  (5. 

IF  4*J.  One  side  of  a  right-angled  triangle,  and  the  sum 
of  the  hypotervu.se  and  the,  otluer  side  being  given,  to  find  the 
hypotenuse  and  the  other  side. 

Ex.  The  sum  of  the  hypotenuse  and  perpendicular  of  a 
right-angled  triangle  is  24  inches,  and  the  base  is  12  inches ; 
what  is  the  length  of  the  hypotenuse,  and  also  of  the  perpen- 
dicular ? 

ANALYSIS.  The  24  inches  is  the  length  of 
the  side  AC  plus  the  length  of  the  side  BC, 
and  consequently  is  the  sum  of  two  num- 
bers. The  12  inches  is  the  length  of  the 
side  AB,  and  is  the  square  root  of  the  differ- 
ence between  the  squares  AC  and  B C.  We 
therefore  have  the  sum  of  two  numbers,  and 
the  square  root  of  the  difference  of  their 
squares,  to  find  the  numbers.  122=144, 

IT  43.    Topic.    Solution  of  Ex.    Rule.    Note. 


52  MENSURATION  OF  LINES  1T  44. 

the  difference  of  the  squares  of  the  two  numbers,  and  144  4-  24  =  6, 
the  difference  of  the  numbers,  (^[  42.)  We  now  have  24,  the  sum  of  two 
numbers,  and  6,  their  difference,  to  find  the  numbers.  (H  41.)  24  +  b 
=  30,  and  30  -f-  2  =  15,  the  greater  number.  24—  6  =  18,  and  18-^-2 

=  9,  the  less  number. 

A        (  Length  of  hypotenuse,      15  inches. 
Ans'    \       «       «  perpendicular,    9     « 

Hence,  When  (me  leg  of  a  right-angled  triangle,  and  the 
sum  of  the  hypotenuse  and  the  other  leg  are  given,  to  find  the 
hypotenuse  and  the  other  leg. 

RULE. 

I.  Divide  the  square  of  the  given  side  by  the  sum  of  the 
other  two  sides,  and  the  quotient  will  be  the  difference  of  the 
two  unknown  sides. 

II.  From  the  sum  and  difference  of  the  two  unknown  sides, 
find  the  two  sides  by  IF  41,  Rule. 

NOTE.    The  operation  may  be  proved  by  If  39,  Rule. 

EXAMPLES  FOR  PRACTICE, 

1.  The  sum  of  the  lengths  of  the  hypotenuse  and  base  of  a  right-angled 
triangle  is  25  feet,  and  the  perpendicular  is  5  feet ;  the  lengths  oMhe 
hypotenuse  and  base  are  required.  An^     j  Hypotenuse,  jo  ft. 

2.  A  gentleman  has  a  triangular  park,  one  of  the  shorter  sides  of 
which  measures  24  rods,  and  the  other  two  sides  measure  72  rods ;  what 
is  the  length  of  each  of  the  other  two  sides  ? 

3.  Two  men.  A  and  B.  started  from  a  certain  place,  and  traveled,  A 
going  east  35  miles,  and  B  north  a  certain  distance,  and  then  in  a  right 
line  to  the  place  at  which  B  had  stopped;  the  distance  that  B  traveled 
was  -jZy  of  the  whole  distance  that  A  and  B  had  both  traveled ;  how  far 
did  if  travel?     How  far  was  he  from  A,  and  also  from  the  place  from 
which  he  started,  when  he  stopped  his  northerly  course  and  turned  to 
meetB?  Answers,  in  order.     49  mi. ;  o7  mi. ;  1^  mi. 

4    A  'tree  100  feet  high  was  broken  off  by  the  wind,  at  such  a  distance 
from  the  bottom,  that  the  top  part  touched  the  ground  50  feet  from  the 
foot  of  the  tree,  the  bottom  of  the  part  broken  off  resting  npon  the  top  o: 
the  stump  ;  what  were  the  hight  of  the  stump,  and  the  length  ot  the  part 
broken  off?  Ans      j  Hight  of  the  stump  37-5  ft. 

Ans"     j  Length  of  part  broken  off,  62-5  " 

IT  44.     The  relation  of  the  three  sides  of  a  right-angled 
triangle  to  each  other,  applied  to  the  measurement  of  distances. 

IT  44.    Topic.     Solution  of  Ex.1. -of  Ex.  2.     Note.     Principles  upon 
which  the  operations  are  performed.    Solution,  of  Ex.  5.  —  of  Ex.  6. 


If  44 


AND  SUPERFICIES. 


53 


1.  In  the  accompanying  figure,  the  base  AB  is  9 
feet  long,  the  perpendicular  BD,  9  feet,  and  the  line 
DE  6  let ;  what  was  the  whole  length  of  the  perpen- 
dicular BC,  of  the  original  triangle  ABC. 

ANALYSIS  It  is  evident  that  the  line  AE,  in  the 
perpendicular  distance  of  9  feet,  inclines  to  or  ap- 
proaches the  perpendicular  B C  3  feet.  The  base  of 
the  triangle  is  9  feet;  and,  in  order  to  complete  t 
triangle,  the  line  AE  must  be  continued  till  it  meets 
the  perpendicular,  that  is,  till  it  has  approached  il 
9  feet.  l  Since  it  approaches  it  3  feet  in  9  it  must  be 
continued  3  times  as  far  to  approach  it  9  feet,  that  is, 
to  meet  it.  3  X  9  feet  =27  feet.  Therefore,  the  per- 
pendicular BC  was  27  feet  long. 

2.  It  is  required  to  find  the 
distance  to  an  inaccessible  point, 
O,  on  the  opposite  side  of  a  river. 

OPERATION.     Place  a  stake  at     ^ -g 

%^^%M'£Z*  that  the  points  A,  B,  and  0,  are  in  the 
same  right  line  AO.  Then,  upon  the  base  AB,  construct  the  square 
A  BCD  Next  from  the  point  D  take  an  observation  ;  note  on  the  line  BC 
the  point  at  which  the  line  DO  will  intersect  the  line  BC,  as  at  B;  and 
meJsnre  the  distance  CE,  say  2  feet.  You  now  have  the  base  AD,U 
feet  •  the  perpendicular  AB,  10  feet ;  and  the  distance  CE,  2  feet ;  to  find 
the  length  of  the  line  AO.  Since  the  line  DE  approaches  the  line  AU  t 
feet  in  10,  in  1  foot  it  approaches  it  TV  of  2  feet,  or  T2(J  =  i  of  a  foot' 

±  ft.  :     1  ft.  :  :   10  ft.  :  50  ft.  ;  or, 

2  ft.   :    10  ft.   :  :   10  ft.   :   50  ft. 

That  is,  the  distance  the  line  DO  approaches  the  line  A 0  in  any  given 
number  of  feet,  is  to  that  number  of  feet,  as  the  whole  line  or  base  AD 
is  to  the  whole  length  of  the  line  AO.  50  feet,  the  distance  from  A  to  0, 
minus  10  feet,  the  distance  from  A  to  B,  leaves  40  feet,  the  distance  Irom 
J5to  O. 

NOTE  The  distance  to  anv  visible  point  on  a  plain,  precipice,  mountain, 
lowa£  &C,  may  be  ascertained,  by  varying  the  foregoing  principles  to  suit  the 
circumstances  uader  which  the  observations  are  taken.  The  observations 
mis  always  be  taken  upon  a  surface  in  the  same  plane  with  the  object, 
unless  the 'elevation  or  depression  of  the  object,  above  or  below  the  plane  on 
which  the  observation  is  taken,  be  known. 

3  Given  the  perpendicular  10  feet,  the  base  10  feet,  and  the  inclina- 
tion'of  the  second  line  of  observation  2£  inches  ;  that  is,  2J  inches  to  10 
feet,  the  length  of  the  perpendicular;    what  is  the  distance  from  the 
nearest  accessible  point  to  the  object  ?  Ans.  4  /  U  J 

4  Wishin«-  to  ascertain  the  distance  to  a  tree  on  the  top  of  a  certain 
hill,'  without  applying  the  chain,  I  constructed  a  square  of  10  feet,  as  in 
the  last  two  examples,  and  found  the  inclination  to  be  J  of  an  inch  ;  wh; 
was  the  distance  to  the  tree  ?  ;  •'•  -   ;  •  -     - 

5* 


54 


MENSURATION  OF  LINES 


1T45 


5.  Wishing  to  know  the  perpendicular 
hight  of  the  hill,  I  constructed  an  instru- 
ment consisting  of  two  legs,  which  re- 
volved  on    a  point,   as   shown    in   the 
accompanying   figure.     The  instrument 
could  be  set  at  any  desirable  angle,  by 

pinning  the  leg  AC  to  the  arc  D.     The 

legs  were  each  10  feet  long.     On  placing 

the  leg  AB  in  a  horizontal  position,  and  elevating  the  end  C  of  the  leg 
AC,  till,  with  my  eye  at  A,  the  upper  edge  of  the  leg  was  in  range  with 
the  foot  of  the  tree,  I  found  the  perpendicular  distance  from  C  to  the  leg 
AS  to  be  just  2  feet ;  allowing  the  leg  AB  to  be  5  feet  above  the  ground, 
what  was  the  perpendicular  hight  of  the  hill?  Aiis,  389  feetr 

6.  On  the  same  hill,  at  the  same  distance  from  me,  was  a  rock,  and  I 
wished  to  ascertain  its  distance  from  the  tree.     To  do  this,  I  used  the 
same  instrument,  with  both  its  legs  lying  in  the  same  plane.     I  placed 
the  point  A  over  the  first  point  of  observation,  the  leg  AB  in  range  with 
the  foot  of  the  tree,  and  the  leg  AC  in  range  with  the  base  of  the  rock. 
Then  the  distance  from  B  to  C  was  4  inches  ;  what  was  the  distance  from 
the  tree  to  the  rock  ?  Am. -  64 -feet. 

IT  4.5.     To  find  the  area  of  a  right-angled  triangle. 

D  C 

ANALYSIS  .  By  an  inspection  of  the  accompa- 
nying diagram,  it  will  readily  be  seen  that  the 
rectangle  ABCD  contains  two  right-angled 
triangles,  ABC  and  ADC ;  consequently  the 
area  of  any  right-angled  triangle  is  equal  to 
one  half  the  area  of  a  rectangle  having  the 
same  base  and  perpendicular.  Hence, 

To  find  the  area  of  a  right-angled  triangle. 

RULE. 

Multiply  one  half  the  base  by  the  perpendicular,  or  one  half 
the  perpendicular  by  the  base.  Or, 

Multiply  the  base  by  the  perpendicular,  and  take  one  half 
of  the  product.* 

NOTEI.  625  sq.  L.,or272'25sq.  ft.,  =  l  P.,  and  16  P.  =  1  A.  (See  If  IT  15 
and  16.)  When  the  area  of  a  triangle  is  in  sq.  L.,  it  will  be  reduced  to  P.  by 

IT  45.    Topic.     Analysis.     Rule.     Note  1.     Reference  to  second 
Ex.  4,  note  3. 

*  The  principles  of  this  rule  may  also  be  ana- 
lyzed as  follows  :  — 

if  from  the  triangle  ABC  a  portion  be  cut  off 
parallel  to  the  base,  as  from  D  to  73,  bisecting1 
the  perpendicular  and  hypotenuse,  the  portion 
CDE  will  be  equal  to  the  triangle  APE,  and,  if 
placed  there,  will  complete  the  rectangle  ABDF, 
which  is  equal  to  the  triangle  ABC. 


IT  46.  AND  SUPERFICIES.  55 


twice  the  number  that  it  takes  of  that  denomination  in  which  the  area  is 
expressed  to  make  one  of  the  next  higher  denomination,  will  reduce  the  area 
to  the  next  higher  denomination.  625  sq.  L.  X  2  =  1250  sq.  L.  ;  272'25  sq. 
ft.  X  2  =  544'5  sq.  ft.  ;  and  16  P.  X  2  =  32  P.  Hence,  To  find  the  number 
of  acres  in  a  triangle,  multiply  the  base  by  the  perpendicular.  If  the  dimen- 
sions are  given  in  links ,  divide  the  product  by  1250,  and  if  in  feet,  by  544 '5, 
and  the  quotient  will  be  poles,  ichich  may  be  reduced  to  acres.  If  the  dimen- 
sions are  given  in  poles,  divide  the  product  by  32,  and  the  quotient  will  be  acres. 
1250  is  $  of  10000,  yV^A  =  £•  Multiplying  by  8  and  dividing  the  product 
by  10000,  is  the  same  in  effect  as  dividing  by  1250.  Therefore,  instead  of 
dividing  by  1250,  we  may  multiply  by  8,  and  cut  off  4  figures  from  the  right 
hand  of  the  product. 

EXAMPLES  FOR  PRACTICE. 

1.  What  are  the  contents  of  a  right-angled  triangle  whose  base  is  12 
inches,  and  perpendicular  8  inches  ? 

2.  The  legs  of  a  right-angled  triangle  are  70  rods,  and  40  rods  m 
length ;  what  are  the  contents  of  the  triangle  in  acres  ?       Ans.  8|  acres. 

3.  The  gable  ends  of  a  barn  are  each  28  feet  wide,  and  the  perpendic- 
ular hight  of  the  ridge  of  the  roof  above  the  eaves  is  7  feet ;  how  many 
feet  of  boards  will  be  required  to  board  up  the  gables  ?        Ans.  196  feet. 

NOTE  2.    By  reference  to  IT  39,  ex.  2,  it  will  be  seen  that  the  gable  may 
readily  be  resolved  into  two  right-angled  triangles. 

4.  The  area  of  a  right-angled  triangle  is  48  feet,  and  the  base  is  12 
feet ;  what  is  the  perpendicular  ? 

NOTE  3.    This  example  is  the  reverse  of  the  preceding  ones  in  this  V,  and 
may  be  performed  by  reversing  the  rule. 

5.  The  area  of  a  right-angled  triangle  is  72  yards,  and  the  perpendicu- 
lar is  9  yards  ;  what  is  the  base  ?  Ans.  16  yards. 

6.  The  area  of  the  gable  of  a  certain  building  is  108  feet,  and  the  per- 
pendicular hight  of  the  ridge  of  the  roof  above  the  eaves  is  9  feet ;  what 
is  the  width  of  the  building  ?  Ans.  24  feet. 

IF  46.     To  find  the  area  of  an  equilateral  and  of  an  isosceles 
triangle. 

ft 

ANALYSIS.  Any  equilateral  triangle,  or  any 
isosceles  triangle,  may  be  resolved  into  two 
right-angled  triangles,  each  having  for  its 
base  one  half  the  base  of  the  given  triangle, 
and  both  having  one  common  perpendicular. 
Thus,  the  equilateral  triangle  ABC  consists 
of  the  two  right-angled  triangles  ADC  and 
BDC.  Hence, 


IT  46.    Topic.    Analysis.    Conclusion,    Note. 


56  MENSURATION  OF  LINES  IT  47. 

The  area  of  an  equilateral,  or  of  an  isosceles  triangle  may  be 
found  by  the  principles  of  the  rule  IF  45. 

EXAMPLES   FOR  PRACTICE. 

1.  What  are  the  contents  of  a  triangle  whose  sides  measure  18  inches 
each?  Ans.  280'44-f- sq.  in. 

2.  What  is  the  area  of  an  equilateral  triangle  whose  side  measures  20 
yards  ?  Ans.  346<4-|-  sq.  yds. 

3.  What  is  the  area  of  an  isosceles  triangle,  the  base  or  longest  side  of 
which  is  16  feet,  and  the  other  sides  are  each  13  feet  4  inches  ? 

Ans.  85  sq.  ft.  4  sq.  in. 

4.  Two  sides  of  a  field,  in  the  form  of  an  isosceles  triangle,  are  each 
60  rods  long,  and  the  other  side  is  96  rods  long ;  how  many  acres  does 
the  field  contain  ?     See  If  45,  Note  1.  Ans.  10  A.  3  R.  8  P. 

NOTE.  When  the  length  of  one  side  of  any  triangle,  and  the  perpendicular 
distance  between  this  side  and  the  opposite  angle,  are  given,  the  area  may  be 
found  by  an  application  of  the  same  principles. 

5.  The  base  of  a  triangle  is  18  inches,  and  the  perpendicular  13  inches ; 
what  is  the  area  ? 

6.  One  side  of  a  field,  in  a  triangular  form,  is  18  chains  in  length,  and 
the  perpendicular  distance  between  this  side  and  the  opposite  angle  is  15 
chains  ;  what  is  the  area  of  the  field  ?  Ans.  3  R.  15  P. 

7.  One  side  of  the  roof  of  a  building  is  16  feet  wide,  the  other  side  18, 
and  the  perpendicular  night  of  the  ridge  above  the  eaves  is  9  feet ;  how 
many  clapboards,  each  covering  4  inches  wide  by  13  feet  long,  will  be 
required  to  cover  both  gables  ?  Ans.  Nearly  60. 

5T  47.     To  find  the  area  of  any  triangle,  the  length  of  the 
sides  being  given 

FIRST  METHOD. 

ANALYSIS.  We  may  con- 
struct any  number  of  trian- 
gles, having  the  same  base 
and  a  common  altitude,  as 
ABC,  ABD,  ABE,  &c., 
and  divide  them  into  right- 
angled  triangles,  by  Tf  35, 
Prob.  VII.,  and  their  areas 
will  all  be  the  same.  Conse- 
quently, all  triangles  con- 
structed on  the  same  base, 
and  having  the  same  alti- 
tude, are  equal.  Hence, 

RULE. 

I.     Construct  the  given  triangle,  as  taught  in  IF  35,  Prob. 
XIV. 

U  47.    Topic.    Analysis  of  first  method.    Rule.    Second  rule.    Note. 


1F48.  AND  SUPERFICIES.  57 

II.  On  one  side  of  the  triangle  erect  a  perpendicular  equal 
to  the  altitude  of  the  triangle. 

III.  Multiply  the  base  by  the  perpendicular,  and  take  one 
half  of  the  product  for  the  area.     See  IF  45. 

SECOND  METHOD. 

RULE. 

I.  Add  the  lengths  of  the  three  sides  together,  and  from 
half  their  sum  subtract  the  length  of  each  side  separately. 

II.  Then  multiply  the  half  length  of  the  three  sides  and  the 
three  remainders  together,  and  extract  the  square  root  of  their 
product. 

NOTE.    The  principles  of  this  rule  do  not  admit  of  an  arithmetical  analysis. 

EXAMPLES  FOR  PRACTICE. 

1.  How  many  poles  in  a  triangular  field,  the  sides  of  which  measure 
respectively  13,  14,  and  15  rods  ?  Ans.  84. 

2.  How  many  square  inches  in  a  triangular  board  whose  sides  measure 
respectively  22,  26,  and  30  inches?  Ans.  278*51-4-. 

3.  A  triangular  field,  whose  sides  measure  386,  420,  and  765  yards 
leases  annually  for  $1<S7A  per  acre  :  how  much  is  the  annual  rent  ? 

Ans.  817'316-f. 

1T  48.     To  find  the  area  of  a  rhombus  and  of  a  rhomboid. 

ANALYSIS.  If  the  right-angled  triangle 
AED  be  placed  on  the  opposite  side  of 
the  rhombus  ABCD,  it  will  fill  the  space 
BFC,  and  the  rhombus  will  then  be  re- 
duced to  a  square.  By  the  same  process 
the  rhomboid  will  be  reduced  to  a  rect- 
angle. Hence, 

A~  E  B      J? 

To  find  the  area  of  a  rhombus  or  of  a  rhomboid. 

RULE. 

Multiply  the  length  by  the  shortest  or  perpendicular  distance 
between  two  opposite  sides. 

EXAMPLES  FOR  PRACTICE. 

1.  The  side  of  a  rhombus  is  18  inches  long,  and  the  shortest  distance 
between  its  opposite  sides  is  14  inches  ;  what  is  its  area? 

2.  A  meadow,  in  the  form  of  a  rhomboid,  is  20  chains  long,  and  the 
shortest  distance  between  its  opposite  sides  is  12  chains  :  how  many  hours 
will  it  take  a  man  to  mow  the  grass  on  this  meadow,  if  he  mow  1  square 
rod  in  3  minutes  ?    How  many  days,  if  he  work  10  hours  each  day  ? 

Ans.  19  da.  2  h. 

IT  48.    Topic.    Analysis.    Rule. 


58  MENSURATION  OF  LINES  T  49,  50. 

3.  The  side  of  a  board,  in  the  form  of  a  rhombus,  is  15  inches  long,  and 
a  perpendicular,  running  from  one  obtuse  angle,  will  meet  the  opposite 
side  9  inches  from  the  acute  angle  •  what  is  the  length  of  the  perpendic- 
ular, and  what  is  the  area  of  the  board  ?  Ans.  to  last.  180  sq.  in. 

^40.     To  find  the  area  of  a  trapezoid. 

The  side  AB  is  24  inches,  the  side 

CD  16  inches,  and  the  altitude  or  dis- 

tance  ad  7  inches  j  what  is  the  area  of 
the  trapezoid  ? 


ANALYSIS.  If  the  triangle  Aae  be 
applied  to  the  space  Dde,  and  the  tri- 
angle  Bbe  to  the  space  Cce,  the  trapezoid  will  oe  reduced  to  a  rectangle 
the  side  ab  being  equal  to  the  side  cd.  The  side  CD  will  be  increased 
just  as  much  as  the  side  AB  is  diminished,  and  the  sides  ab  and  cd  will 
each  be  equal  to  one  half  the  sum  of  the  sides  AB  CD  24  4-  16  = 
40,  40  H-  2  =  20,  and  20  X  7  =  140,  the  number  of  square  inchesin  the 
trapezoid.  Hence, 

To  find  the  area  of  a  trapezoid. 
RULE. 

Multiply  one  half  the  sum  of  the  parallel  sides  by  the  per- 
pendicular distance  between  them. 

EXAMPLES  FOR  PRACTICE. 

1.  What  are  the  square  contents  of  a  board  12  feet  long,  16  inches 
wide  at  one  end,  and  9  at  the  other  ?  Ans.  12£  sq.  ft. 

2.  What  are  the  contents  of  a  stock  of  12  boards,  14  feet  long,  10 
inches  wide  ai  one  ena.  and  2£  at  the  other  ?  Ans.  87£  sq.  ft. 

3.  What  is  the  area  of  a  board  12  feet  long,  16  inches  wide  at  each 
end,  and  8  in  the  middle  ?  Ans.  12  sq.  ft. 

4.  One  side  of  a  field  is  40  chains  long,  the  side  parallel  to  it  is  22 
chains  long,  and  the  perpendicular  distance  between  these  two  sides  is  25 
chains  ;  how  many  acres  in  the  field  ?  Ans.  77  A.  5  sq.  C. 

*![  «5O.     To  find  the  area  of  a  trapezium. 

ANALYSIS.  If  from  any  angle  a  diagonal 
be  drawn  to  the  opposite  angle,  as  AC,  the 
figure  will  be  divided  into  two  triangles, 
ABC  and  A  CD.  Consequently,  if  the  sides 
and  a  diagonal  of  a  trapezium  be  given,  the 
sides  of  the  triangles  which  compose  it  are 
also  given.  Hence, 


IT  49.    Topic.    Analysis.     Rule. 
IT  50.    Topic.    Analysis.    Rule. 


1T51.  AND  SUPERFICIES.  59 

To  find  the  area  of  a  trapezium. 

RULE. 

I.  Divide  the  trapezium  into  two  triangles,  by  drawing  a 
diagonal  to  two  opposite  angles,  and  from  each  of  the  other 
two  angles  let  fall  a  perpendicular  to  the  diagonal. 

II.  Multiply  the  sum  of  the  perpendiculars  by  one  half  the 
diagonal,  or  the  diagonal  by  one  half  the  sum  of  the  perpen- 
diculars.    Or, 

Find  the  area  of  each  triangle  separately,  IF  47,  and  add 
them  together. 

EXAMPLES  FOR  PRACTICE. 

1.  The  diagonal  of  a  trapezium  is  25  feet,  and  the  perpendiculars  are 
9  and  13  feet ;  what  is  the  area  ?  Ans.  275  sq.  ft. 

2.  The  diagonal  of  a  trapezium  is  34  feet  9  inches,  and  the  sum  of  the 
perpendiculars  is  28  ft.  6  inches  ;  what  is  the  area  ?      Ans.  495  ft.  2'  3". 

3.  The  diagonal  of  a  trapezium  is  32  yards,  the  two  remaining  sides 
of  one  triangle  are  20  and  26  yards,  and  the  two  remaining  sides  of  the 
other  are  29  and  9  yards  ;  what  is  the  area  of  the  trapezium  ? 

Ans.  387<65-f-  sq.  yds. 

4.  The  diagonal  of  a  field,  in  the  form  of  a  trapezium,  is  538  yards,  the 
two  remaining  sides  of  one  triangle  are  283  and  471  yards,  and  of  the 
other  432  and  216  yards  ;  how  many  acres  in  the  field? 

Ans.  22  A.  3  R.  31  P.  18<66  sq.  yds. 

^T  51.     Similar  rectilinear  figures. 

1.  What  ratio  does  a  figure  4  inches  square  bear  to  one  2  inches 
square?     ANALYSIS.  42=lo,  2*=  4,  and  16-^-4  =  4. 

Ans.     The  ratio  of  4  to  1. 

2.  How  does  the  area  of  a  square  foot  compare  with  that  of  a  square 
yard*  Ans.     It  is  £  as  large. 

3.  I  have  4  small  fields ;  the  largest  is  40  rods  square,  and  the  other  3 
are  each  20  rods  square ;  how  does  the  largest  field  compare  in  size  with 
one  of  the  others  ?     How  with  all  of  them  ?  Ans.  to  last.     As  4  to  3. 

4.  How  many  boards,  3  feet  long  and  2  feet  wide,  will  be  required  to 
cover  a  space  9  feet  lung  and  6  feet  wide  ? 

5.  Two  rectangles  measure  as  follows  :  the  smaller  5  by  12  inches,  and 
the  larger  20  by  48  inches ;  how  many  of  the  smaller  ones  will  be  equal 
to  the  larger  one  ?  Ans.  16. 

6.  The  two  shorter  sides  of  one  right-angled  triangle  measure  8  and  15 
feet,  and  the  two  shorter  sides  of  another  16  and  30  feet ;  how  do  the 
areas  compare  with  each  other2  Ans.  As  1  to  4. 

7.  The  3  sides  of  one  triangle  measure  respectively  18,  14,  and  10 
inches,  and  the  3  sides  of  another  9,  7,  and  5  inches ;  what  ratio  does  the 
area  of  the  first  bear  to  that  of  the  second  ?          Ans.  The  ratio  of  4  to  1. 

8.  The  side  of  a  regular  octagon  measures  9  inches,  and  the  side  of  a 

If  51.    Topic.    Object  of  Ex.  1.— of  Ex.  2. —  Ex.  4.— Ex.  C.  — Ex.  8. 


60  MENSURATION  OP  LINES  1T  52,  53. 

second  similar  figure  measures  45  inches ;  the  second  is  how  many  times 
as  large  as  the  first  ?    How  many  times  larger  ? 

Ans.  to  last.     24  times  larger. 

Ex.  1  shows  that  a  square  whose  sides  are  double  the 
length  of  another  square,  contains  4  times  the  area. 

Ex.  2  shows  that  a  square  whose  side  is  3  times  as  long  as 
the  side  of  another  square,  contains  9  times  the  area. 

Ex.  4  shows  that  a  rectangle  twice  as  long  and  twice  as 
wide  as  another  rectangle,  contains  4  times  the  area. 

Ex.  6  shows  that  any  triangle  whose  sides  measure  twice  as 
much  as  the  sides  of  another  similar  triangle,  contains  4  times 
the  area. 

Ex.  8  shows  that  any  regular  polygon  whose  side  is  5  times 
as  long  as  the  side  of  another  similar  polygon,  contains  25 
times  the  area. 

In  each  of  the  preceding  examples  in  this  IT,  the  area  of  the 
larger  figure  can  be  obtained,  by  multiplying  the  area  of  the 
smaller  figure  by  the  square  of  the  number  of  times  the  side 
of  the  smaller  figure  is  contained  in  the  side  of  the  larger,  and 
vice  versa.  Hence, 

IT  52.  The  areas  of  similar  rectilinear  figures  are  to  each 
other  as  the  squares  of  their  similar  sides  ;  and 

The  similar  sides  are  to  each  other  as  the  square  root  of  the 
quotient  of  the  area  of  the  greater  figure  divided  by  the  area  of 
the  less. 

NOTE.  The  pupil  will  find  by  trial  that  the  second  of  the  above  principles 
is  the  reverse  01  the  first ;  and  that,  having  the  areas  of  two  similar  figures, 
and  any  side  of  one  of  them  given,  the  similar  side  of  the  other,  and  the  ratio 
between  all  the  similar  sides  of  the  two,  may  be  found,  by  an  application  of 
these  principles. 

IT  58.     To  find  the  area  of  any  regular  polygon. 

ANALYSIS.  Any  regular  polygon  may  be  re- 
solved into  as  many  equal  triangles  as  the  polygon 
contains  sides.  The  base  of  each  triangle  will 
be  the  length  of  one  side  of  the  polygon,  and  the 
altitude  of  each  will  be  the  perpendicular  distance 
from  the  center  of  the  polygon  to  the  middle  of 
one  side.  Hence, 


IT  52.    First  principle.     Second.     Note. 
IT  53.    Topic.    Analysis.    Rule.    Note. 


IT  54. 


AND  SUPERFICIES. 


61 


To  find,  the  area  of  any  regular  polygon. 

RUL.E. 

I.  Bisect  any  two  contiguous  sides  of  the  polygon ,   the 
point  where  the   bisecting  lines  intersect  each  other  will  be 
the  center  of  the  polygon,  or  the  apex  of  the  triangle  of  which 
the  polygon  is  composed ;  and  the  perpendicular  distance  from 
the  center  to  one  side  will  be  the  altitude  of  each  triangle. 

II.  Multiply  the  length  of  the  perpendicular  by  one  half 
the  perimeter. 

NOTE.  We  may  first  find  the  area  of  one  triangle,  IT  4.7,  and  then  multiply 
this  area  by  the  number  of  triangles  in  the  given  polygon. 

EXAMPLES  FOR  PRACTICE. 

1.  "What  is  the  area  of  a  regular  pentagon,  the  side  of  which  measures 
25  inches,  and  the  perpendicular  distance  from  the  center  to  the  middle 
of  one  side  is  17'2  inches?  Ans.  7  sq.  ft.  67  sq.  in. 

2.  What  is  the  area  of  a  regular  hexagon,  the  side  of  which  is  10  feet, 
and  the  altitude  of  one  of  its  equal  triangles  is  8<660254  feet  ? 

Ans.  259<80762  sq.  ft. 

^T  54.  The  areas  of  similar  polygons  are  to  each  other  as 
the  squares  of  one  of  their  sides.  IF  52.  Hence,  the  areas  of 
regular  polygons  may  be  more  readily  found  by  the  help  of  a 
table  prepared  in  the  following  manner. 

Consider  the  sides  of  each  of  the  regular  polygons  to  be  1 ; 
then  find  the  perpendicular  and  area. 


Number  of  sides. 

Names  of  figures. 

Areas  and  Multipliers. 

3 

Triangle, 

'433013 

4 

Square, 

I'OOOOOO 

5 

Pentagon, 

1  '720  177 

6 

Hexagon, 

2<59S076 

7 

Heptagon, 

3<633912 

8 

Octagon, 

4<828427 

9 

Nonagou, 

6<181824 

10 

Decagon, 

7<694209 

11 

Undecagon, 

9<365640 

12 

Dodecagon, 

U<196152 

The  above  table  shows  the  area  of  polygons  of  any  number 
of  sides  from  3  to  12,  each  side  being  unity  or  1.  Hence, 

The  length  of  one  side,  and  the  number  of  sides  of  any  regu- 
lar polygon  being  given,  to  find  the  area,  by  the  above  table. 


TT  5-1.    Principle.    Reference.    Formation  of  table.    Its  use.    Rule.    Note. 

6 


62  MENSURATION  OF  LINES  If  55. 

RULE. 

Multiply  the  square  of  the  given  side  by  the  tabular  number 
of  a  similar  polygon. 

NOTE  If  the  area,  the  number  of  sides,  and  the  length  of  one  side  be 
given  the  perpendicular  may  be  found  by  reversing  this  rule.  Also,  if  the 
frea  the  number  of  sides,  and  the  perpendicular  be  given  the  length  of  one 
side  may  be  found  by  reversing  the  rule.  Also  the  area  the  perpendicular 
and  the  length  of  one  side  being  given,  the  number  of  sides  may  be  found  by 
a  reverse  process. 

EXAMPLES  FOR  PRACTICE. 

1   What  is  the  area  of  a  triangle  whose  side  is  15  inches  ? 

Ans.   97'4-f-  sq.  in. 
2.  What  is  the  area  of  a  square  whose  side  is  5  miles  ? 

3  What  is  the  area  of  a  pentagon  whose  side  is  8  feet  ? 

Ans.    110  sq.ft.  15'9-f-  sq.  in. 

4  What  is  the  area  of  a  hexagon  whose  side  is  2$  yards  ? 

Ans.  16  sq.  yds.  2  sq.  ft.  20'4-f-  sq.  in. 

5.  What  is  the  area  of  a  heptagon  whose  side  is  £  of  a  rod  ? 

Ans.  6  sq.  yds.  7  sq.  ft.  120  sq.  in. 

6    The  side  of  an  octagon  is  4£  feet ;  what  is  the  area? 

Ans.   90'667-j-  sq.  ft. 

7.  The  side  of  a  nonagon  is  11$  inches  ;  what  is  the  area  ? 

Ans.    12'897+  sq.  ft. 

8.  The  side  of  decagon  is  3|  yards ;  what  is  the  area  ? 

Ans.     78  sq.  yds.  7  sq.  ft.  14+  sq.  in. 

Q    The  side  of  an  undecagon  is  46  chains  and  15  links ;  what  is  the 
area?  Ans.  1994  A.  7-15-1-  sq.  C. 

10      The  side  of  a  dodecagon  is  $  of  an  inch ;  what  is  the  area  ? 

Ans.  1'57-J-sq.  in. 

IT  55.  To  fold  the  area  of  any  irregular  rectilinear  fig- 
ure or  polygon. 

RULE. 

I.  Divide  the  figure  into  as  many  triangles  as  may  be,  by 
drawing  diagonals  from  any  one  angle  to  all  the  others. 

II.  Find  the  areas  of  the  several  triangles,  and  add  them 
together. 

NOTE  1 .  Any  rectilinear  figure  may  be  divided  into  as  many  triangles,  less 
two,  without  any  of  the  dividing  lines  crossing  each  other,  as  the  figure  has 

NOTE  2.  The  area  may  frequently  be  more  easily  found,  by  dividing  the 
polygon  into  as  many  trapezia  as  may  be,  thus  diminishing  the  number  ot 
triangles  ;  for  the  area  of  a  trapezium  is  more  readily  obtained  than  the  areas 
of  the  two  triangles  which  compose  it. 

""  IT  55.    Topic.    Rule.    Note  1.    Note  2. 


1f  56, 57.  AND  SUPERFICIES.  63 

EXAMPLES  FOR  PRACTICE. 

1.  A  piece  of  land  of  an  irregular  figure,  bounded  by  right  lines,  is 
divided  into  3  trapezia:  in  the  first  the  diagonal  is  4  C.  24  1.,  and  me 
sum  of  the  perpendiculars  3  C.  67  1. ;  in  the  second  the  diagonal  is  7  b. 
43  L,  and  the  sum  of  the  perpendiculars  5  C.  38  1 . ;  and  in  the^ third  the 
diagonal  is  6  C.  78  1.,  and  the  sum  of  the  perpendiculars  4  C.  84 :  I. ;  how 
many  acres  does  the  field  contain?  4"§-  4' 


BOARD  OR  LUMBER  MEASURE. 

If  56.  The  standard  of  thickness  for  boards  is  1  inch. 
All  lumber  not  exceeding  1  inch  in  thickness  is  bought  and 
sold  by  the  superficial  measure  of  144  square  inches  to  the 
foot.  All  lumber  exceeding  1  inch  in  thickness  is  first  redu< 
to  the  standard  thickness,  and  then  estimated  by  superficial 
measure,  as  before. 

IF  57.     To  find  the  number  of  feet  in  a  straight-edged 
board,  of  uniform  width. 

For  rule  and  principles  consult  H  37. 

To  find  the  number  of  feet  in  a  board  that  tapers. 

For  rule  and  principles  consult  TT  49. 

NOTE  1.  The  mean  width  of  a  tapering  board  is  usually  taken,  by  meas- 
uring the  width  of  the  board  at  an  equal  distance  from  each  end. 

EXAMPLES  FOR  PRACTICE. 

1.  How  many  feet  in  a  board  58  inches  long  and  9  inches  wide  ? 

NOTE  2.  When  the  length  and  width  are  given  in  inches,  divide  the  prod- 
uct by  144  to  reduce  to  square  feet.  Why? 

2.  How  many  feet  in  a  board  18  feet  long  and  15  inches  wide  ? 

NOTE  3.  When  the  length  is  given  in  feet,  and  the  width  in  inches,  divide 
the  product  by  12  to  reduce  to  square  feet.  Why  1 

3.  How  many  feet  in  4  boards,  each  12  ft.  7  in.  long,  and  1  ft.  4  in. 
wide  ? 

NOTE  4  The  inches  may  be  reduced  to  fractions  of  a  foot,  and  the  whole 
to  improper  fractions,  and  multiplied  ;  the  feet  may  be  reduced  to  inches,  and 
then  multiplied  ;  or,  the  multiplication  may  be  performed  by  duodecimals. 
See  Revised  Arithmetic,  11204. 

IT  56.  Standard  thickness  of  boards.  Lumber  less  than  1  inch  thick. 
Lumber  more  than  1  inch  thick. 

IT  57.  Number  of  feet  in  a  straight-edged  board  of  uniform  width;  —  in  a 
straight-edged  board  that  tapers.  Mean  width  of  a  tapering  board.  Both  di- 
mensions in  feet ;  —  in  inches.  One  dimension  in  feet  and  the  other  in  inches. 
Dimensions  in  feet  and  inches. 


64  MENSURATION  OF  LINES  IF  58,  59. 

4.  How  many  feet  in  a  stock  of  9  boards,  each  13  feet  long,  9  inches 
wide,  and  U  inches  thick  ?  Af- . l ^ *  • 

5.  How  many  feet  in  a  stock  of  5  boards,  each  11  feet  long,  15  inches 
wide  at  one  end,  and  12  inches  at  the  other  ? 

6.  How  many  feet  in  a  stock  of  13  boards,  each  14  feet  long,  16  inches 
wide  at  one  end,  and  running  to  a  point  at  the  other  ?  Ans.  Wl|, 

7    What  length  of  board  9  inches  wide  will  make  a  square  toot  t 

8.  What  length  of  board  15  inches  wide  will  be  required  to  make  a 
fireboard  3  feet  square  ? 

9.  How  many  feet,  board  measure,  in  a  stock  of  7  planks,  each  12  feet 
long,  22  inches  wide,  and  2\  inches  thick  ?  *£  •    *?*' 

10.  What  are  the  contents  of  a  stock  of  9  boards,  13  feet  long,  18  inches 
wide  at  each  end,  and  14  in  the  middle  ? 

11    What  are  the  contents  of  a  stock  of  7  boards,  13  feet  long, ,10 
inches  wide  at  one  end,  12  at  the  other,  and  8  at  the  distance  of  t 
from  the  wider  end  ?  4iw.  ™  sq.  it.  48  sq.  in. 

IT  58.  To  find  the  number  of  feet  of  straight-edged  boards 
in  a  stock  of  wane-edged  boards,  sawn  from  a  round  log,  without 

Blabbing.  ,  ,     -  - 

In  sawing  a  round  log  into  wane-edged  boards,  a  slab  ol  1 
inch  is  taken  off  from  two  opposite  sides,  and  the  remainder 
of  the  log  is  sawn  into  boards,  which  of  course  are  a*  differ- 
ent widths.  From  an  actual  measurement  of  stocks  of  boards, 
sawn  from  logs  in  the  manner  above  described,  the  following 
facts  have  been  deduced.  . 

1.  In  stocks  of  boards  sawn  from  logs  from  7  to  12  inches  in 
diameter,  the  width  of  the  second  board  is  the  average  width,  oj 
the  whole  stock.  . 

2  In  stocks  of  boards  sawn  from  logs  from  12  to  24  inc/ies  in 
diameter,  the  width  of  the  third  board  is  the  average  width  of 
the  whole  stock.  «••'_» 

3.  In  stocks  of  boards  sawn  from  logs  from  24  to  36  inches  in 
diameter,  the  width  of  the  fourth  board  is  the  average  width  of 
the  whole  stock.  _. 

In  all  cases  a  slab  of  1  inch  in  thickness  is  to  be  taken  ott, 
before  the  first  board  is  sawn;  and  the  board  is  to  be  meas- 
ured on  the  narrower  side;  i.  e.,  the  side  towards  the  slab. 
The  above  averages  will  vary  slightly,  but  will  not  make  a  dil- 
ference  of  more  than  2J  feet  in  any  log  from  7  to  3»>  mdu>- : 
the  difference  being  sometimes  in  favor  of  the  buyer,  and  at 
others  in  favor  of  the  seller. 

IT  59.     The  diameter  of  a  circle  being  given,  to  find  the 

circumference.  _ 

IT  58.  Sawing  wane-edged  boards.  First  fact.  Second.  Third.  Slab- 
bing. Measuring  the  average  board.  Variation  of  these  averages. 


IT  59.  AND  SUPERFICIES.  65 

4*,.' 

It  has  been  a  problem  among  the  ablest  mathematicians,  for 
ages,  to  find  the  exact  ratio  between  the  diameter  and  circum- 
ference of  a  circle.  Although  this  ratio  has  never  yet  been 
definitely  ascertained,  yet  results  have  been  arrived  at,  which 
are  sufficiently  exact  for  all  practical  purposes. 

It  has  been  found  that  the  diameter  of  a  circle  is  to  the  circum- 
ference nearly  as  1  to  3f ,  or  7  to  22.  A  nearer  approximation 
is  as  113  to  355,  or  1  to  344159.  The  former  ratio  is  suffi- 
ciently exact  for  ordinary  mechanical  purposes,  but  in  esti- 
mating machinery,  and  other  calculations  where  greater  accu- 
racy is  required,  the  latter  ratio  should  be  used.  Hence, 

When  the  diameter  of  a  circle  is  given,  to  find  the  circum- 
ference. 

RULE. 

Multiply  the  diameter  by  3f ;  or,  where  greater  accuracy 
is  required,  by  3'14159.  Or,  for  ordinary  purposes,  say, 

7  I  22  : ;  the  given  diameter  :  the  required  circumference. 
And,  where  greater  accuracy  is  required,  113  :  355  ::  the 
diameter  :  the  circumference. 

NOTE  1.  The  latter  ratio  has  the  advantage  of  being  easily  remembered, 
the  numbers  being  formed  of  the  first  three  odd  numbers,  each  repeated. 

NOTE  2.  Most  of  the  operations  under  this  head,  in  this  work,  are  per- 
formed by  the  proportion  113  :  355  :  :  the  given  diameter  :  the  circumference. 

EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  circumference  of  a  circle  8  feet  in  diameter  ? 

Ans.  25ft.  1'58-f  in. 

2.  "What  is  the  circumference  of  a  circle  7  inches  in  diameter,  by  the 
first  proportion ? — by  the  second?    Which  gives  the  greater  result,  and 
how  much  ? 

Ans.  The  first  gives  '00885  of  an  inch  greater  than  the  second. 

3.  The  diameter  of  a  certain  wheel  is  10'5  feet ;  what  is  its  circum- 
ference? Ans.  32  ft.  11'83-j-in. 

4.  What  length  of  tire  will  it  take  to  band  a  carriage- wheel  5  feet  in 
diameter  ? 

5.  What  is  the  circumference  of  a  circle  113  rods  in  diameter,  by  the 
first  proportion  ?  —  by  the  second  ?    Which  gives  the  greater  result,  and 
how  much  ? 

6.  What  is  the  circumference  of  a  circular  lake  721  rods  in  diameter? 

Ans.  7  mi.  25  rds.  1'454-ft. 

7.  A  horse  is  made  fast  to  a  stake  by  a  line,  one  end  of  which  is 
fastened  to  his  nose,  and  the  other  to  the  stake  ;  allowing  this  line  to  be 
25  feet  long,  what  is  the  circumference  of  the  circle  upon  which  he  may 
feed?  Ans.  157-07-j-it. 

1159.    Topic.    Analysis.    Rule.    Note  1.    Note  2. 
6* 


6&  MENSURATION  OF  LINES  H  60,  61. 

IT  6O.     The  circumference  of  a  circle  being  given,  to  find 
the  diameter. 

RULE. 

Divide  the  circumference  by  3|;  or,  where  greater  accuracy 
is  required,  by  3'  14159. 

Or,  for  ordinary  purposes,  say  22  :  7  :  :  circumference  ! 
diameter.     And  where  greater  accuracy  is  required,  355  :  113 
:  ;  circumference  :  diameter. 

NOTE.     Since  this  rule  is  the  reverse  of  the  rule  IT  59,  no  analysis  is  deemed 
necessary. 

EXAMPLES  FOR  PRACTICE. 

1  What  is  the  diameter  of  a  circle  33  yards  in  circumference  ? 

2  If  the  circumference  be  49<52  rods,  what  is  the  diameter  ? 

Ans.  15'762-|-  rds. 

3.  The  circumference  of  a  cart-wheel  is  16  feet  6  inches  :  what  is  the 
diameter  ? 

4.  A  circular  park  is  320  links  in  circumference  ;  what  is  its  diameter? 

5.  If  the  extreme  end  of  the  minute-hand  of  a  clock  move  forward  19 
inches  in  12  minutes,  what  is  the  circumference  of  the  dial-plate  ?    AVhat 
is  the  length  of  the  minute-hand  ?  Ans.  to  last.  15  j1^  inches. 

6.  Within  a  circular  garden  66  chains  in  circumference,  is  a  circular 
pond  66  rods  in  circumference  ;  what  is  the  diameter  of  the  garden  ?  — 
of  the  pond  ,  Ms,        Dieter  of 


IT  61.     The  number  of  degrees  in  a  circular  arc,  and  the 
radius  of  the  circle  being  given,  to  find  the  length  of  the  arc. 

RULE. 

I.  Find  the  circumference  of  the  circle. 

II.  Then  say,  360°  :  the  number  of  °  in  the  arc  :  :  the  cir- 
cumference of  the  circle  :  the  length  of  the  arc. 

NOTE.    The  reasons  for  the  rule  are  obvious. 

EXAMPLES  FOR  PRACTICE. 

1    What  is  the  length  of  an  arc  of  18°,  in  a  circle  whose  radius  is  4  ft. 

g,5  jn  ?  Ans.  1  ft.  5<75  in. 

"2.  What  is  the  length  of  an  arc  of  36°  15',  in  a  circle  whose  radius  is 

153few?hat  is  the  length  of  an  arc  of  225°  7'  30",  in  a  circle  12  rods  in 

diameter?  ^ins.  7  rds.  8  ft.  3'72-j-in. 

"4    The  length  of  an  arc  is  17|  inches,  and  the  remaining  part  of  the 

circle  contains  342°  j  what  is  the  diameter  of  the  circ1^   2Q  ft  ?  .Q 

IT  60.    Topic.    Analysis.    Rule. 
IT  61.    Topic.    Analysis.    Rule. 


1162,63.  AND  SUPERFICIES.  67 

IT  62.     To  find  the  area  of  a  circle. 

ANALYSIS.  Any  circle  may  be  supposed  to  be  divided  into  an  infinite 
number  of  equal  isosceles  triangles,  whose  apexes  all  meet  in  the  center 
of  the  circle,  and  whose  bases  all  lie  in  the  circumference.  In  other 
words,  the  circle  may  be  considered  as  a  regular  polygon  of  an  infinite 
number  of  sides,  the  perimeter  of  the  polygon  being  the  circumference 
of  the  circle,  and  the  altitude  of  one  of  the  equal  triangles  of  which  it  is 
composed  being  the  radius.  And,  since  the  area  of  a  regular  polygon  is 
found  by  multiplying  the  perpendicular  by  one  half  the  perimeter,  (IT  53, 
rule,)  the  area  of  a  circle  may  be  found  by  applying  the  same  principles. 
Hence, 

To  find  the  area  of  a  circle. 

RULE. 

Multiply  one  half  the  circumference  by  one  half  the  diam- 
eter. 

NOTE  1.  If  the  whole  circumference  be  multiplied  by  the  whole  diameter, 
and  the  product  divided  by  4,  the  quotient  will  be  the  area.  This  operation, 
will  in  many  cases  obviate  the  use  of  fractions  in  multiplying  and  dividing. 

EXAMPLES  FOR  PRACTICE. 

1.  The  diameter  of  a  circle  is  7,  and  the  circumference  22  j  what  is 
the  area  ? 

2.  The  diameter  of  a  circle  is  113,  and  the  circumference  355 ;  what  is 
the  area  ? 

3.  What  is  the  area  of  a  barrel  head  16  inches  in  diameter? 

NOTE  2.    First  find  the  circumference,  by  IT  59,  rule. 

Ans.   201<04  sq.  in. 

4.  What  is  the  area  of  a  circle  described  with  a  radius  2  chains  in 
length  ? 

5.  The  circumference  of  the  end  of  a  log  is  82  inches ;  what  is  the 
area  ?  Ans.  3  sq.  ft.  lOS'OS-j-sq.  in. 

^T  63.  To  find  the  area  of  a  circle,  when  the  diameter  only 
is  given. 

Ex.  What  is  the  area  of  a  circle  inscribed  within  a  figure  1 
foot  square  ? 

ANALYSIS.  The  diameter  of  the  circle  is  equal  to  one  side  of  the 
square,  or  1  foot.  1  ft.  (diameter)  ^-  2  =  '5  foot ;  344159  ft.  (circum- 
ference) -f-  2  ==  1 '570795  ft.  Then,  '5  X  1 '570795  =  '7853975,  the  area 
of  any  circle  whose  diameter  is  1.  This  area  wants  but  '0000025  = 
TWb'o'TT  °^  being  '7854,  which  is  so  small  a  fraction,  that  in  business 
calculations  it  is  disregarded,  and  the  area  of  a  circle  is  estimated  to  be 
'7854  of  the  area  of  its  superscribing  square.  Hence, 

IT  62.    Topic.    Analysis.    Rule.  Note  1. 

IT  63.    Topic.    Solution  of  Ex.    Conclusion  deduced  from  solution.    Rule. 


68  MENSURATION  OP  LINES  1F  64, 6-5. 

To  find  the  area  of  a  circle,  when  the  diameter  only  is  given. 

RULE. 

Multiply  the  square  of  the  diameter  by  '7854. 

EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  area  of  a  circle  7  inches  in  diameter? 

2.  What  is  the  area  of  one  surface  of  a  circular  saw  25  inches  in  diam- 
eter ?  Ans.  3  sq.  ft.  58J  sq.  in. 

3.  What  is  the  area  of  a  circle  |  of  an  inch  in  diameter  ? 

4.  How  many  circles  1  inch  in  diameter,  are  equal  to  a  circle  4  inches 
in  diameter  ?  -A»«-  16- 

IT  64.     The  area  of  a  circle  being  given,  to  find  the  diameter. 

RULE. 

Divide  the  area  by  '7854,  and  extract  the  square  root  of  the 
quotient. 

NOTE  1 .  This  rule  being  the  reverse  of  the  rule  1T63,  an  analysis  is  deemed 
unnecessary. 

EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  diameter  of  a  circle  whose  area  is  38£  square  feet  ? 

Ans.  1  ft.  '01-j-in. 

2.  What  is  the  diameter  of  a  circular  park  which  contains  2464  square 
yards  ? 

3.  The  area  of  a  circle  is  78<54  chains;  what  is  the  circumference  ? 

NOTE  2.    First  find  the  diameter. 

Ans.  310.41*59+1. 

4.  What  is  the  diameter  of  a  circular  island  containing  1  square  mile 
of  land?  Ans.  1  mi.  41  rds.  5'83-f- ft. 

5.  What  is  the  circumference  of  a  circular  pond  which  covers  7<06S6 
square  chains  ? 

IT  65.  To  find  the  area  of  a  semicircle,  a  quadrant,  and  a 
sextant. 

ANALYSIS.  The  area  of  a  semicircle  is  equal  to  £,  the  area  of  a  quad- 
rant to  £,  and  the  area  of  a  sextant  to  £  the  area  of  a  circle  having  the 
same  radius.  Hence, 

To  find  the  area  of  a  semicircle,  a  quadrant  and  a  sextant. 

IT  64.    Topic.     Analysis.     Rule. 

IT  65.  Topic.  Analysis.  Rule  for  the  semicircle  ;  —  for  the  quadrant ; 
for  the  sextant.  Note  1.  Reasons  for  these  operations. 


IF  66.  AND  SUPERFICIES.  69 

RULE. 

I.  For  the,  semicircle;  —  Take  J  of  the  area  of  a  circle 
having  the  same  radius. 

II.  For  the  quadrant;  —  Take  J  the  area  of  a  circle  having 
the  same  radius. 

III.  For  the  sextant ; — Take  £  the  area  of  a  circle  having 
the  same  radius. 

NOTE  1 .  The  area  of  the  semicircle  may  be  found,  by  multiplying  £  the 
square  of  twice  the  radius  by  '7854,  or  twice  the  square  of  the  radius  by  '7854. 
Also,  the  area  of  the  quadrant  may  be  found,  by  multiplying  i  the  square  of 
twice  the  radius  by  '7854,  or  the  square  of  the  radius  by  '7854.  The  intelligent 
pupil  will  readily  deduce  these  principles  from  IT  63,  rule,  and  from  the  prin- 
ciples of  simple  multiplication. 

EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  area  of  £  of  a  circle  16  inches  in  diameter? 

2.  What  is  the  area  of  a  semicircle  whose  radius  is  18  feet  ? 

3.  What  is  the  area  of  a  quadrant  whose  radius  is  4  yards  ? 

4.  What  is  the  area  of  a  sextant  whose  radius  is  11  inches  ? 

5.  What  is  the  radius  of  a  quadrant  whose  area  is  12 '5664  square 
yards  ?  Ans.  12  ft. 

NOTE  2.  This  example  is  the  reverse  of  Ex.  3.  The  pupil  will  deduce  his 
principles  for  operation  from  IT  45,  Note  3. 

6.  The  area  of  a  semicircle  is  3<5343  square  chains  j   what  is  the 
radius  ?  Ans.  18  rds. 

7.  The  area  of  a  sextant  is  63*3556  square  inches  ;  what  is  the  radius  ? 

5T  G6.  To  find  the  area  of  a  sector,  the  radius  and  arc 
being  given. 

ANALYSIS.  Since  the  whole  circumference  of  a  circle  contains  the 
whole  area,  any  arc  and  its  radii  will  contain  such  a  part  of  the  whole 
area,  as  the  arc  is  part  of  the  whole  circumference.  Or, 

A  sector  may  be  supposed  to  consist  of  an  infinite  number  of  triangles, 
the  same  as  the  circle  in  ^f  62.  Hence, 

To  find  the  area  of  a  sector,  the  radius  and  arc  being  given. 

RULE. 

First,  find  the  circumference  of  the  original  circle,  and  also 
its  area.  Then  say, 

The  whole  circumference  :  the  given  arc  : :  the  whole  area 

the  area  of  the  sector.      Or, 

Multiply  the  length  of  the  arc  by  J  the  radius. 

EXAMPLES  FOR  PRACTICE. 

1.  The  radius  of  a  sector  is  15  inches,  and  the  arc  4£  inches  j  what  is 
the  area? 

TT6G.    Topic.    Analysis.    Rule. 


70  MENSURATION  OF  LINES  IT  67,  68. 

2.  The  radius  is  48  yards,  and  the  arc  50  yards  ;  what  is  the  area  of 
the  sector  ? 

3.  What  is  the  area  of  a  sector  whose  radius  is  200  rods,  and  arc  12A 
feet? 

4.  The  area  of  a  sector  is  33|  inches,  and  the  radius  15  inches  j  what 
is  the  arc  ? 

NOTE.    This  example  is  the  reverse  of  Ex.  1. 

^F  O71.  To  find  the  area  of  a  sector,  the  radius  and  the  angle 
at  the  center  being  given. 

ANALYSIS.  Since  the  whole  circumference  of  a  circle  contains  360°, 
the  sum  of  all  the  angles  that  can  be  made,  by  radii  drawn  from  the  cir- 
cumference to  the  center  of  any  circle,  must  be  360° ;  and  any  sector 
must  contain  such  a  part  of  the  area  of  the  whole  circle,  as  the  number 
of  degrees  contained  in  its  angle  at  the  center  is  part  of  360°.  Hence, 

To  find  the  area  of  a  sector ',  the  radius  and  angle  at  the  center 
being  given. 

RULE. 

First  find  the  area  of  the  original  circle.     Then  say, 
360°  :  the  given  angle  : ;  the  area  of  the  original  circle  : 
the  area  of  the  sector. 

EXAMPLES  FOR  PRACTICE. 

1.  The  radius  of  a  sector  is  7  feet,  and  the  angle  at  the  center  45° ; 
what  is  the  area  of  the  sector  ?  Ans.  19<2423  sq.  ft. 

2.  The  radius  of  a  sector  is  24  ft.  6  in.,  and  the  arc  contains  137°  30' ; 
what  is  the  area  of  the  sector  ?  Ans.  196  sq.  ft.  62<181  sq.  in. 

3.  The  radius  of  a  sector  is  113  inches,  and  the  area  17  sq.  ft.  59<19315 
sq.  in. ;  what  is  the  angle  at  the  center  ?  Ans.  22°  30'. 

NOTE.    This  example  is  the  reverse  of  the  two  preceding. 

IF  68.  To  find  the  side  of  a  square  which  shall  contain  an 
area  equal  to  a  given  circle. 

RULE. 

Find  the  area  of  the  given  circle,  and  extract  its  square 
root. 

NOTE.    The  pupil  will  readily  analyze  this  rule. 

EXAMPLES  FOR  PRACTICE. 

1.  "What  is  the  side  of  a  square  equal  in  area  to  a  circle  14  feet  in 
diameter?  Ans.  12'407-f- ft. 

IT  67.    Topic.    Analysis.     Rule, 
1T68»    Topic.    Analysis.    Rule. 


H69.  AND  SUPERFICIES.  71 

2.  I  have  a  circular  garden  42  rods  in  circumference,  and  I  wish  to  lay 
out  a  square  park  of  the  same  area  as  the  garden ;  what  will  be  the  length 
of  one  side  of  the  park  ? 

3.  One  monument  is  built  upon  a  circular  base,  and  another,  which 
stands  near  it,  upon  a  square  base,  the  area  of  the  base  of  the  latter 
being  equal  to  that  of  the  former.      The  side  of  the  base  of  the  square 
monument  is  5  feet  j  what  is  the  diameter  of  the  base  of  the  circular  mon- 
ument? Ans.  5'641-f-ft 

^T  G9.  The  diameter  of  a  circle  being  given,  to  find  the  side 
of  its  inscribed  equilateral  triangle. 

ANALYSIS.  The  side  AB,  of  the  equilateral 
triangle  ABC,  intersects  the  radius  DE  of  the 
superscribing  circle,  at  the  point  d,  equi-distant 
from  D  and  E.  Consequently,  the  side  AB 
is  the  base  of  two  triangles,  AdE  and  BdE, 
the  hypotenuse  of  each  being  the  radius  of  the 
circle,  or  the  side  of  the  inscribed  hexagon,  and 
the  perpendicular  of  each  £  the  radius  DE. 
We  now  have  the  hypotenuse  AE  or  BE, 
(equal  to  the  radius  of  the  given  circle,)  and 
the  perpendicular  Ed,  (equal  to  £  the  radius,) 
to  find  the  base  Ad  or  Bd,  which,  being  multiplied  by  2,  will  be  the  side 
of  the  inscribed  equilateral  triangle.  Hence, 

To  find  the  side  of  an  equilateral  triangle  inscribed  in  a 
given  circle. 

RULE. 

I.  From  the  square  of  the  radius  subtract  the  square  of 
half  the  radius. 

II.  Extract  the  square  root  of  the  remainder,  and  multiply 
the  result  by  2. 

NOTE  1.  Instead  of  the  second  operation  directed  in  the  rule,  the  same 
result  will  be  obtained  by  multiplying  the  remainder  by  4,  and  extracting  the 
square  root  of  this  product. 

EXAMPLES  FOR  PRACTICE. 

1.  The  radius  of  a  circle  is  8  inches  •  what  is  the  side  of  an  inscribed 
equilateral  triangle  ?  Ans.  13'85-{-  in. 

2.  What  is  the  side  of  an  equilateral  triangle  inscribed  in  a  circle  50 
yards  in  diameter  ?  Ans.  86  yds.  1  ft.  9-69-j-  in. 

3.  What  is  the  side  of  the  greatest  equilateral  triangle  that  can  be  cut 
from  a  circular  plate  of  copper  3  inches  in  diameter  ?        Ans.  2'598-{-  in. 

4.  The  side  of  an  equilateral  triangle  is  14  inches ;  what  is  the  diam- 
eter of  the  superscribing  circle  ? 

NOTE  2.     This  example  is  the  reverse  of  the  three  preceding  ones. 

Ans.  16'16-f-in. 

IT  69.    Topic.    Analysis.    Rule.    Note  1.    Method  of  performing  Ex.  4. 


72  MENSURATION  OF  LINES  IF  70. 

^T  TO.     The  diameter  of  a  circle  being  given,  to  find  the 
side  of  its  inscribed  square. 

D 


ANALYSIS.  The  diameter  of  the  circle  is  the 
diagonal  of  the  square  ABCD,  and  is  also  the 
hypotenuse  of  the  two  right-angled  triangles 
ABCzn&CDA.  Hence, 


To  find  the  side  of  a  square  inscribed  in  a  given  circle. 

RULE. 

Square  the  diameter,  divide  the  square  by  2,  and  extract 
the  square  root  of  the  quotient.  Or, 

Square  the  radius,  multiply  the  square  by  2,  and  extract  the 
square  root  of  the  product. 

NOTE  1.  The  pupil  will  readily  perceive,  upon  an  examination  of  this 
rule,  that  the  area  of  a  square  inscribed  in  a  circle  is  equal  to  £  the  square  of 
the  diameter,  or  twice  the  square  of  the  radius  of  the  circle. 

EXAMPLES  FOR  PRACTICE. 

1.  The  diameter  of  a  circle  is  10  inches ;  what  is  the  side  of  its  in- 
scribed square  ?  Ans.  7'07-j-  in. 

2.  What  will  be  the  side  of  a  stick  of  square  timber  hewn  from  a  log 
2  feet  in  diameter  ? 

3.  The  diameter  of  a  circle  is  3  yards  ;  what  is  the  area  of  its  inscribed 
square  ?  Ans.  4£  sq.  yds. 

4.  The  circumference  of  a  circle  is  11  inches  ;  what  is  the  side  of  its 
inscribed  square  ? 

NOTE  2.     First  find  the  diameter  of  the  circle. 

5.  The  side  of  a  square  is  4  feet ;  what  is  the  diameter  of  its  super- 
scribing circle  ? 

6.  The  area  of  a  square  is  49  inches ;  what  is  the  radius  of  its  super- 
scribing circle  ?  Ans.  4'94-j-  in. 

U  70.    Topic.    Analysis.    First  rule.    Second.    Note  1. 


11  71,  72.  AND  SUPERFICIES.  73 

*fl"  71*     The  diameter  of  a  circle  being  given,  to  find  the 
side  of  its  inscribed  octagon. 

ANALYSIS.  If  from  the  radius  HO,  we  sub- 
tract'the  line  KO,  (equal  to  £  the  side  of  the 
inscribed  square,)  the  remainder  HK  will  be 
the  perpendicular,  and  the  line  AK  or  GK, 
(equal  to  £  the  side  of  the  inscribed  square,) 
will  be  the  base  of  the  right-angled  triangle, 
AKH  OY  GKH ;  and  the  hypotenuse  ^.Hor 
GIfjVfill  be  one  side  of  an  octagon  inscribed 
in  the  same  circle.  Hence, 

To  find  the  side  of  an  octagon  inscribed  in  a  given  circle. 

RULE. 

I.  Find  the  side  of  the  inscribed  square,  by  IF  70,  rule,  and 
subtract  \  of  it  from  the  radius  of  the  circle. 

II.  Square  this  remainder,   and  also   J  the  side    of  the 
inscribed  square. 

III.  Add  the  squares  together,  and  extract  the  square  root 
of  their  sum. 

NOTE.    The  side  of  a  hexagon  inscribed  in  a  circle  is  equal  to  the  radius 
of  the  circle. 

EXAMPLES  FOR  PRACTICE. 

1.  The  radius  of  a  circle  is  5  inches ;  what  is  the  side  of  its  inscribed 
octagon  ?  Ans.  3'82-f-  in. 

2.  What  will  be  the  side  of  an  octagon  inscribed  in  a  circle  24  inches 
in  diameter? 

3.  A  gentleman  laid  out  a  garden  in  the  form  of  an  octagon,  the  radius 
of  whose  circumscribing  circle  was  7  poles  ;  what  was  the  length  of  one 
side  of  the  garden  ?  Ans.  5  rds.  5  ft.  10'82-j-  in. 

4.  The  circumference  of  a  circle  is  44  inches-  what  is  the  side  of  its 
inscribed  hexagon  ? 

5.  The  side  of  a  hexagon  inscribed  in  a  circle  is  7  inches  ;  what  is  the 
side  of  a  square,  arid  also  of  an  octagon,  inscribed  in  the  same  circle  ? 

5F  72*     The  two  axes  of  an  ellipse  being  given,  to  find  the 
area. 

RULE. 

Multiply  the  two  diameters  together,  and  their  product  hy 

'7854.       ' 

NOTE  1.     This  rule  being  deduced  from  the  principles  given  in  TTG3,  a  repe- 
tition of  those  principles  in  this  place  is  deemed  unnecessary. 

Tf  71.    Topic.     Analysis.     Rule.     Note. 
H  72.    Topic.    Analysis.    Rule.    Note  2. 

7 


•74  MENSURATION  OF  LINES 

EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  area  of  an  ellipse  whose  axes  are  15 
>.  The  transverse  diameter  of  an  ellipse  is  30  rods, 


73,74. 


'  diameter 

100  feet  j  what  is  the  conjugate  diameter  ? 

NOTE  2.    When  the  area  and  one  axis  are  given,  the  other  axis  may  be  found 
by  dividing  the  area  by  '7854,  and  that  quotient  by  the  glven  axis.^    ^  ^ 

4  The  area  of  an  ellipse  is  '7854  of  a  foot,  and  one  axis  is  9  inches  ; 
what  is  the  other  axis? 

IT  73.  To  find  the  diameter  of  a  circle  whose  area  shall  le 
equal  to  the  area  of  a  given  ellipse. 

RULE. 

Multiply  the  axes  of  the  given  ellipse  together,  and  extract 
the  square  root  of  the  product. 

NOTE  The  same  result  will  be  obtained,  by  finding  the  area  of  the  given 
ellipsl,  and  from  that  area  finding  the  diameter  of  the  required  circle. 

For  analysis  of  principles  consult  HIT  63  and  64. 

EXAMPLES  FOR  PRACTICE. 

1.  The  axes  of  an  ellipse  are  35  and  48  feet  ;  what  is  the  diameter  oi 

30 
is  339^2928  square  feet  •  what  is 


of  a  circle  of  equal  area? 

<T  74  To  find  the  area  of  the  space  contained  between  the 
arcs  of  four  equal  circles,  adjacent  to  each  other,  and  all 
lying  at  an  equal  distance  around  a  point. 


ANALYSIS.  The  square  AB  CD 
contains  ^  of  the  area  of  each  circle, 
and  also  the  space  contained  by  the 
arcs  EF,  FG,  GH,  and  HE.  %  of 
the  area  of  one  circle  multiplied  by 
4,  is  equal  to  one  whole  circle ;  and 
the  area  of  the  square,  whose  side  is 
twice  the  radius  of  one  circle,  minus 
the  area  of  one  circle,  is  the  area  of 
the  required  space.  Hence, 


IT  73.    Topic.     Analysis.    Rule.     Note. 
IT  74.    Topic.     Analysis.    Rule. 


IT  75.  AND  SUPERFICIES.  75 

To  find  the  area  of  the  space  contained  between  the  arcs  of  4 
equal  circles,  adjacent  to  each  other,  and  all  lying  at  an  equal 
distance  around  a  point. 

RULE. 

Subtract  the  area  of  one  circle  from  the  square  of  twice  the 
radius. 

EXAMPLES  FOR  PRACTICE. 

1.  The  radius  of  each  of  4  equal  circles,  lying  at  an  equal  distance  from 
a  point,  is  9  inches  ;  what  is  the  area  of  the  space  contained  between  the 
arcs  of  the  circles?  Ans.  69'53-{-  sq.  in. 

2.  The  diameter  of  each  of  4  equal  circles,  lying  adjacent  to  each  other, 
at  an  equal  distance  about  a  point,  is  11  inches;  what  is  the  area  of  the 
space  contained  between  the  arcs  of  the  circles  ? 

3.  What  is  the  area  of  the  space  contained  between  the  arcs  of  4  equal 
circles,  situated  as  in  the  last  example,  the  area  of  one  circle  being  26<18 
square  chains  ?  Ans.  7'153-[-  sq.  C. 

*[T  7«i.  To  find,  the  area  of  the  space  contained  between  the 
arcs  of  three  equal  circles,  adjacent  to  each  other,  and  all  lying 
at  an  equal  distance  around  a  point. 

ANALYSIS.  The  equilateral 
triangle  ABC  contains  •£  of  the 
area  of  each  circle,  and  also  the 
area  of  the  space  contained  be- 
tween the  arcs  DE,  EF,  and  FD. 
£  of  the  area  of  3  equal  circles 
is  equal  to  f  or  J-  the  area  of  one 
of  them  ;  and  the  area  of  the 
equilateral  triangle,  whose  side 
is  twice  the  radius  of  one  circle, 
minus  3-  the  area  of  one  circle, 
is  the  area  of  the  required  space. 
Hence, 

To  find  the  area  of  the  space  contained  between  the  arcs  of 
three  equal  circles,  adjacent  to  each  other,  and  all  lying  at  an 
equal  distance  around  a  point. 

RULE. 

Subtract  J  the  area  of  one  circle  from  the  area  of  an  equi- 
lateral triangle  whose  side  is  twice  the  radius  of  one  circle. 

EXAMPLES  FOR  PRACTICE. 

1.  The  radius  of  each  of  3  adjacent  circles,  lying  at  an  equal  distance 
about  a  point,  is  2£  feet;  what  is  the  area  of  the  space  contained  between 
their  arcs?     '  Ans.   1 ''00875  sq.  ft. 

2.  The  diameter  of  each  of  3  circles,  situated  as  in  the  last  example,  is 
2g-  yards  ;  what  is  the  area  of  the  space  contained  between  their  arcs  ? 

IT  75.    Topic.    Analysis.    Rule. 


76  MENSURATION  OF  LINES,  ETC.  1T  76, 77. 

3.  The  circumference  of  each  of  three  circles  situated  as  in  the  last  two 
examples,  is  71  chains  j  what  is  the  area  of  the  space  contained  between 
their  arcs  ?  Ans.  20<566-f-  sq.  C. 

1T  76.     To  find  the  area  of  a  circular  ring. 


ANALYSIS.  A  circular  ring  is  the  space 
included  between  the  circumferences  of  two 
concentric  circles  of  different  diameters. 

Its  area  must  evidently  be  equal  to  the 
area  of  the  larger  circle  minus  the  area  of 
the  smaller.  Hence, 


To  find  the  area  of  a  circular  ring. 

RULE. 

Square  the  two  diameters,  subtract  the  less  square  from  the 
greater,  and  multiply  the  remainder  by  '7854.  Or, 

Multiply  the  sum  of  the  two  diameters  by  their  difference, 
and  this  product  by  '7854. 

NOTE.  The  result  in  either  case  is  the  same  as  would  be  obtained  by  sub- 
tracting the  area  of  the  less  circle  from  the  area  of  the  greater.  The  product 
ot  the  sum  oi  the  diameters  multiplied  by  their  difference,  is  equal  to  the  dif- 
ference of  their  squares.  (See  IT  42.)  ' 

EXAMPLES  FOR  PRACTICE. 

1.  Within  a  circular  park  15  rods  in  diameter,  is  a  circular  pond  7  rods 
in  diameter ;  what  is  the  area  of  that  part  of  the  park  not  covered  bv  the 
pond? 

2.  In  a  pleasure-ground  is  a  circular  pond,  in  the  middle  of  which  is  a 
circular  island  ;  the  diameter  of  the  pond  is  100  yards,  and  the  circum- 
ference of  the  island  the  same  j  what  is  the  area  of  the  pond  ? 

Ans.  7058<22-f-  sq.  yds. 

3.  A  farmer  has  an  elliptical  orchard,  whose  axes  are  300  and  200 
yards,  and  he  wishes  to  surround  it  with  a  wall  3  feet  thick  within  the 
boundary  line  ;  how  much  land  will  be  covered  by  the  wall  ? 

Ans.  25  P.  26-}-  sq.  yds. 

^T  77,     Similar  curmlijiear  figures. 

The  areas  of  similar  curvilinear  figures,  as  of  similar  recti- 
linear figures,  are  to  each  other  as  the  squares  of  their  similar 
radii,  diameters,  circumferences,  curves,  or  linear  dimensions ; 
and  the  similar  radii,  diameters,  circumferences,  curves,  or 

IT  76.  Topic.  Analysis.  Rule.  Note.  Reasons  for  the  operations 
directed  in  the  note.  Method  of  performing  Ex.  3. 

IT  77.  Topic.  Principles.  Where  first  demonstrated  in  this  work.  To 
what  applied  in  that  IT. 


W 

Tf  78.  PRACTICAL  EXAMPLES. 

linear  dimensions  are  to  each  other  as  the  square  root  of  the 
quotient  of  the  area  of  the  greater  figure  divided  by  the  area 
of  the  less.  See  IT  52. 

EXAMPLES  FOR  PRACTICE. 

1.  What  must  be  the  diameter  of  a  circle,  to  contain  4  times  the  area 
of  a  circle  2  inches  in  diameter  ? 

2  The  circumference  of  a  circle  is  38  inches ;  what  must  be  the  cir- 
cumference of  a  circle  containing  16  times  the  area  ?  A^m^r 

3.  The  diameter  of  a  circle  is  15  inches  ;  what  must  be  the  diamete 
of  that  circle  whose  area  is  £  as  great  ? 

4.  The  radius  of  one  circle  is  4  inches,  and  of  another  9^  what^  w  the 

ra5°  The^adiuTof  one  sector  is  2  feet,  and  of  another  similar  one  6  j 

W6\l The  arc^of^wo^imiiar  quadrants  are  10  and  20  inches;  what  is 
the  ratio  of  their  radii,  and  of  their  areas  ? 

7  The  transverse  diameter  of  an  ellipse  is  9  yards  ;  what  must  be  the 
transverse  diameter  of  a  similar  ellipse,  whose  area  is  25  times  , 


Ans.  45  yds. 


PRACTICAL    EXAMPLES 

IN  THE  MENSURATION  OF  LINES  AND  SUPERFICIES. 

IT  78.   1    A  mason  plastered  a  room  22  ft.  long,  19  ft.  9  in.  wide,  and 

9  ft.  6  in.  high  -,  how  much  did  he  receive  for  the  job,  at  $  18  a  square 

i  p  Ans.  -p-i  I'OOij. 

2.  How  much  must  be  paid  for  glazing  3  windows,  each  3<5  ft.  1 by 
7<75  ft.,  at  10  cents  a  square  foot  ? 

3  A  building  30  feet  high  stands  on  the  bank  of  a  stream  50  feet  wide ; 
what  is  the  length  of  a  ladder  that  will  reach  from  the  opposite  bank  of 
the  stream  to  the  top  of  the  building  ? 

4  The  base  of  a  right-angled  triangle  is  62  feet,  and  the  area  1210 
feet';  what  is  the  length  of  the  perpendicular  ?  —  of  the  hypotenuse  ? 

Ans.  Perpendicular  40  ft.     Hypotenuse  bb-bb-f  ft. 

5.  The  perpendicular  is  9,  and  the  sum  of  the  base  and  hypotenuse  19  ; 

what  is  the  base  ?  —  the  hypotenuse  ?  .    .     (  Base,  7TV 

'    I  Hypotenuse,  11-J.f . 

6  The  base  is  22  inches,  and  the  sum  of  the  perpendicular  and  hypot- 
enuse 44  inches  :  what  is  the  area  ?  Ans.  II  1-5  sq  in. 

7.  The  width  of  a  certain  barn  is  21  feet,  and  the  length  of  the  ralte 
13  feet  •  what  will  be  the  length  of  a  purline  beam  extending  from  any 
rafter  to  the  opposite  one,  and  meeting  the  rafters  at  the  distant 

feet  from  the  bottom  of  each  ?  Ans-  UTS  ft; 

8.  A  field  GO  rods  long  contains  15  acres  ;  how  many  rods  in  length  of 
the  same  field  will  be  required  for  9  acres? 

9  I  have  a  triangular  board  containing  94  square  inches,  the  base,  or 
longest  side,  being  234  inches  in  length.  I  wish  to  divide  this  board  into  -> 
parts  each  having  the  same  altitude  as  the  whole  triangle,  and  containing 
respectively  25, 33.  and  30  square  inches  ;  what  will  be  the  lengthof  the  base 
of  each  piece  ?  (See  T[  47.)  Answers,  in  order.  6-'25  in. ;  8(25  in. ;  9  in. 
7* 


73  PRACTICAL  EXAMPLES.  IT  78. 

10.  The  three  sides  of  a  triangular  park  measure  respectively  20,  29, 
and  30  chains  ;  how  many  acres  in  the  park  ? 

Ans.  27  A.  7  sq.  C.  2  P.  268  sq.  ft.  47'46-fsq.  in. 

11.  "What  is  the  area  of  an  equilateral  triangle,  whose  side  measures 
44  inches?  Ans.  838'3-4-sq.  in. 

12.  A  certain  rectangular  field  contains  82  A.  5  P.  of  land,  and  its 
length  is  to  its  breadth  as  7  is  to  3  ;  what  are  the  dimensions  of  the  field  ? 

Ans.  75  and  175  rods. 

13.  "What  length  of  a  mahogany  plank,  26  inches  wide,  will  make  1£ 
square  yards?  Ans.  6'23-j-ft. 

14.  A  triangular  field,  whose  legs  measure  900  and  1775  links,  rents 
for  $37<50  per  annum  ;  how  much  is  that  an  acre  ?  Ans.  $4'694-|-. 

15.  There  is  a  house  three  stories  high,  with  7  windows  in  each  story. 
Each  window  is  2  ft.  8  in.  wide,  and  the  hight  of  the  windows  in  the 
first  story  is  6  ft.  10  in.,  in  the  second  story  5  ft.  8  in.,  and  in  the  third 
story  5  ft.  6  in.  •  what  will  the  glazing  come  to,  at  $'14  a  square  foot  ? 

Ans.  §47'04. 

16.  "What  will  the  paving  of  a  rectangular  court-yard  come  to,  at  $-'75 
a  square  yard,  the  yard  being  42  ft.  9  in.  front,  by  68  ft.  6  in.  deep? 

Ans.  $244<03-J-. 

17.  The  roof  of  a  house  is  52  ft.  8  in.  long,  and  45  ft.  9  in.  from  one 
eave  to  the  other,  across  the  ridge  ;  what  will  the  roofing  cost,  at  $2'25 
a  square  ?  Ans.  $54(21-{-. 

18.  I  have  a  stick  of  timber  3£  by  8£  inches,  and  I  want  another  stick 
just  twice  as  large,  and  4^  inches  thick ;  what  must  be  its  width  ? 

Ans.  13'22-f-in. 

19.  A  wheelwright  made  a  carriage-wheel  4  ft.  10  in.  in  diameter,  and 
the  rim  consisted  of  7  fellies ;  what  was  the  length  of  each  felly  ? 

Ans.  2  ft.  2'03-f-in. 

20.  The  areas  of  2  similar  parallelograms  are  to  etich  other  as  9  to  7£, 
and  the  shorter  side  of  the  smaller  parallelogram  is  19  rods ;  what  is  the 
length  of  the  shorter  side  of  the  larger  parallelogram  ?     Ans.  20'81-|-  rods. 

21.  The  radius  of  a  quadrant  is  21  inches  ;  what  is  the  arc  of  another 
quadrant,  whose  area  is  £  as  great  as  the  former?  Ans.  19(04-j-in. 

22.  What  is  the  diameter  of  that  circle  whose  area  is  12  times  as  great 
as  that  of  a  circle  25  inches  in  diameter?  Ann.  86'6-[-iri. 

23.  A  pillar  7  inches  in   diameter  is  sumcient   to  sustain  a  certain 
weight;  what  must  be  the  diameter  of  a  pillar  that  shall  sustain  10  times 
the  weight,  the  length  of  the  2  pillars  being  the  same  ?   Ans.  22'13-j-in. 

24.  Three  pipes,  each  3  inches  bore,  will  fill  a  reservoir  in  a  certain 
lime  ;  what  must  be  the  diameter  of  the  bore  of  a  pipe  that  will  fill  a  reser- 
voir 2£  times  as  large  in  the  same  time?  Ans.  8-21-j-in. 

25.  What  is  the  diameter  of  a  circular  pond  that  covers  £  acre  of  sur- 
face? Ans.  10  rds.  1  ft.  6'3-4-in. 

26.  What  is  the  length  of  a  cord,  one  end  of  which  being  fastened  to  a 
stake,  and  the  other  end  to  a  horse's  nose,  will  permit  the  horse  to  graze 
upon  a  semicircle  containing  just  1  acre  of  ground  ?   Ans.  10  rds.  1'525-j-ft. 

27.  There  is  a  room  16  feet  long,  15  feet  wide,  and  9  feet  high  ;  what 
is  the  nearest  distance  from  any  corner  at  the  bottom  to  the  diagonal 
corner  at  the  top?  Ans.  23;706-{-ft. 

28.  A  painter  engaged  to  paint  a  church  86  feet  long,  50  feet  wide,  20 
feet  high  to  the  top  of  the  beams,  and  17  feet  from  the  beams  to  the  ridge, 
for  $'37£  per  square  yard  ;  how  much  would  the  job  come  to,  no  deduc- 
tions being  made  for  windows,  doors,  &c.,  nor  no  additions  for  mould- 
ings, cornices,  &c.  ?  Ans.  $262(08-f. 


IT  78. 


PRACTICAL  EXAMPLES.  79 

29.  A  portion  of  railroad  1  mile  in  length  passes  through  the  farms 
of  3  menf  as  follows  :  70  rods  through  the  first  farm,  115  ^ds  through 
the  second,  and  the  remainder  through  the  third.     The  owner  of  the  first 
farm  was  awarded  $83<50  per  acre  as  damages  the  owner  o  Qthe  second 
farm  S92<37£  per  acre,  and  the  owner  of  the  third  farm  J&10C  P6'  acre. 
Allowing  the  road  to  be  4  rods  wide,  how  ^h  did  each  """J^gg* 

Ans.  First  man,  ei46'12i  ;  second,  $265<5  /8+  5  third  & 

30.  The  shadow  of  a  staff  3  feet  long,  at  a  certain  hour  of  the  day 
measures  4  ft.  8  in.  ;  what  is  the  hight  of  that  tree  whose  shadow  at  the 
same  time  measures  179  ft.  5  in.  ?  •  4  T\  1J1 

31.  The  wheels  of  a  rail-car  are  each  2  ft  8  in   in  diameter: 
many  revolutions  do  they  make  in  a  minute,  when  the  cars  are  junmng 


(\C\  Yw1«  1  AflS.     I   A.  O  Slj.   \J-    1U  J.   J.  • 

34  In  a  river  are  6  circular  islands,  the  diameters  of  5  of  which  are 
10<5,  16*  20'25,  26<75,  and  32  rods,  respectively.  The  area  of  the  sixth  is 
equal  to'  the  sum  of  the  areas  of  all  the  others  ;  what  is^its 


A  B  C  and  D  bought  a  grindstone  40  inches  in  diameter,  for 
&ey  paid  $6^0,  eacgh  paying  an  equal  share.  A  first  used  the 
stone  t  11  he  had  ground  off  his  share  ;  B  then  did  the  same,  and  so  with 
C  and  D  Whatgwas  the  diameter  of  the  stone  when  it  came  into  the 
hand»of-B,C,andD,j^v^  34(64+in  ;  _. 

36    A  gentleman  laid  out  a  circular  pleasure-ground,  which  cont 
28  P  231  sq   ft  of  land.     He  then  laid  out  a  graveled  walk  on  the  outer 


in?-machine.  The  sweep  is  12  feet  long,  .he  horse=  i  are  3  feet  apart  and 
tlie  evener  upon  which  llK-y  work  is  attached  to  the  end  of  »«sw«^j 
SuDDOsine  the  horses  lo  make  2  circuits  m  a  minute,  and  to  work  8  lull 
"?y  how  much  farther  will  the  horse  working  upon  the  outs.de 
.  the  day,  than  the  one  working  upon  to  inside^ 


39.  AD  and  BC  are  the  fronts  of 
two  houses,  standing  on  opposite 
sides  of  a  public  square  ;  and  EF  is 
a  post  standing  in  the  square,  in  a 
right  line  between  the  houses.  The 
hight  of  AD  is  55  feet,  and  of  BC 
64  feet.  The  distance  from  the  foot 
of  the  post  to  the  base  of  the  house 
BC,  is  76  feet;  from  the  top  of  the 
post  to  the  top  of  the  house  BC,  9o 


80  MENSURATION  OF  SOLIDS.  IF  79. 

feet ;  and  from  the  top  of  the  post  to  the  top  of  the  house  A  Z>,  80  feet. 
What  is  the  hight  of  the  post  ?  —  the  distance  from  the  base  of  one  house 
to  the  base  of  the  other  ?  —  from  the  top  of  one  house  to  the  top  of  the 
other?  Answers,  in  order.  7  ft. ;  140  ft.  j  140<2S-f-ft. 


MENSURATION  OF  SOLIDS. 

IT  79.     To  find  the  cubic  contents  of  a  prism,  cube,  parallel- 
opiped,  cylinder,  or  cylindroid. 

RULE. 

Multiply  the  area  of  one  end  by  the  length,  or  the  area  of 
the  base  by  the  altitude. 

NOTE  1.     For  analysis  of  principles,  see  Revised  Arithmetic,  IT  51. 
EXAMPLES  FOR  PRACTICE. 

1.  The  side  of  a  cubic  block  measures  8  inches ;  how  many  cubic 
inches  does  it  contain? 

2.  How  many  cubic  feet  in  a  cube  whose  side  measures  11  feet  ? 

3.  The  end  of  a  square  prism  is  10  inches  square,  and  the  length  is  2£ 
feet  =  30  inches  ;  how  many  cubic  feet  does  it  contain  ? 

NOTE  2.  When  the  three  dimensions  are  in  inches,  divide  the  cubic  con- 
tents by  1728  ;  when  iwo  dimensions  are  in  inches  and  the  third  in  feet,  divide 
by  144  ;  and  when  two  dimensions  are  in  feet  and  the  third  in  inches,  divide 
by  12  ;  and  iu  either  case  the  quotient  will  be  cubic  feet.  Why  ? 

4.  The  end  of  a  prism  20  feet  long  is  a  right-angled  triangle,  the  two 
shorter  sides  of  which  measure  9  and  12  inches ;  what  are  the  cubic  con- 
tents of  the  prism  ? 

5.  What  are  the  contents  of  a  parallelepiped  15  feet  long,  3  feet  wide, 
and  11  inches  thick?  Ans.  41$  cu.  ft. 

6.  "What  is  the  solidity  of  a  cylinder  7  feet  long,  and  2  feet  in  diameter  ? 

7.  What  are  the  contents  of  a  log  31  feet  long,  and  17£  inches  in 
diameter?  Ans.  51 '78-}- cu.  ft. 

8.  What  are  the  solid  contents  of  a  stick  of  timber  28  feet  long,  and  8 
inches  square  ?  Ans.  12  cu.  ft.  7o'S  cu.  in. 

9.  A  stick  of  timber  is  25  ft.  3  in.  long,  1  ft.  8  in.  wide,  and  18  in. 
thick ;  how  much  will  it  come  to,  at  8  cents  per  cubic  foot? 

NOTE  3.     Consult  IT  57,  Note  4.  Ans.  $5'05. 

10.  What  is  the  solidity  of  a  block  of  marble  10  ft.  long,  5$  ft.  wide, 
and  3£  ft.  thick  ? 

11.  A  cistern  is  5£  feet  in  diameter,  and  8  feet  deep;  how  many 
standard  gallons  will  it  contain  ?  Ans.  1421<79S4  gal. 

12.  The  diameters  of  a  cylindroidal  tube  200  feet  long,  are  3  and  5 
inches  ;  how  many  standard  gallons  will  it  contain  ?         Ans.  122;4  gals. 

13.  The  side  of  the  base  of  a  regular  hexagonal  prism  is  9  inches,  and 
the  altitude  is  14  feet ;  what  is  the  solidity  ? 

IT  79.    Topic.    Rule.    Analysis.    Note  2.    Note  4. 


80. 


MENSURATION  OF  SOLIDS. 


81 


14.  What  is  the  solidity  of  a  regular  octagonal  pillar  26  feet  long,  one 
side  of  which  measures  7  inches?    (See  If  54.)       Ans.  42'718-j-cu.  ft. 

NOTE  4.  The  superficial  contents  of  any  of  the  figures  named  in  this  IT 
may  be  obtained,  by  multiplying  the  circumference  or  the  girth  of  one  end  of 
the'figure  by  the  length,  and  to  the  product  adding  the  areas  of  the  two  ends. 
Why  ? 

15.  What  is  the  surface  of  a  cube  whose  side  is  4  feet  ? 

16.  The  end  of  a  prism  25  feet  long  is  an  equilateral  triangle,  the  side 
of  which  measures  16  inches  ;  what  is  the  area  of  the  prism  ? 

Ans.  10 1'539-f- sq.ft. 

17.  What  is  the  surface  of  a  prism  18  feet  long,  and  21  inches  square? 

18.  What  are  the  superficial  contents  of  a  round  pillar  14  inches  in, 
diameter,  and  30  feet  long?  Ans.  102  sq.  ft.  20  sq.  in. 

IT  80.     To  find  the  cubic  contents  of  a  pyramid  or  a  cone. 


ANALYSIS.  The  cubic  contents  of  any  pyramid, 
of  a  given  base  and  altitude,  are  equal  to  J  of  the 
cubic  contents  of  a  prism  having  the  same  base 
and  altitude.  And, 


The  cubic  contents  of  any  cone,  of  a  given  base 
and  altitude,  are  equal  to  J  of  the  cubic  contents 
of  a  cylinder  having  the  same  base  and  altitude. 
Hence, 


To  find  the  cubic  contents  of  a  pyramid  or  a  cone. 

RULE. 

Multiply  the  area  of  the  base  by  \  of  the  altitude ;  or, 
Multiply  the  area  of  the  base  by  the  altitude,  and  take  J  of 
the  product 
NOTE  1.    The  correctness  of  this  rule  may  be  verified  by  rule  IT  84. 

EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  solidity  of  a  pyramid  15  feet  square  at  the  base,  and  40 
feet  high? 

2.  Each  side  of  the  base  of  a  triangular  pyramid  is  30  inches,  and  the 
altitude  is  4  feet;  what  are  the  cubic  contents?  Ans.  10:825  cu.  ft. 

IT  80.    Topic.    Analysis.    Rule.     Proof.    Note  2.    Note  3.     Note  4. 


82  MENSURATION  OP  SOLIDS.  1T  81. 

3.  The  area  of  the  base  of  an  octagonal  pyramid  is  78  square  feet,  and 
the  altitude  is  19£  feet ;  what  is  the  solidity? 

4.  The  base  of  a  cone  is  7  feet  in  diameter,  and  the  altitude  is  16  feet 
9  inches ;  what  are  the  solid  contents  ? 

5.  The  altitude  of  a  cone  is  5  feet,  and  the  circumference  of  the  base  5\^ 
feet;  what  are  the  cubic  contents?  Ans.  4'64295-j-cu.  ft. 

6.  The  slant  hight  of  a  cone  is  18  inches,  and  the  diameter  of  the  base 
15  inches  ;  what  is  the  solidity  ? 

NOTE  2.  The  slant  hight  of  a  cone  is  the  distance  from  the  vertex  to  tb 
circumference  of  the  base,  and  the  slant  hight  of  a  pyramid  is  the  distanc 
from  the  vertex  to  the  middle  of  one  side  of  the  base. 

7.  What  is  the  solidity  of  a  pyramid  30  feet  square  at  the  base,  the 
slant  hight  being  25  feet  ?  Ans.  6000  cu.  ft. 

NOTE  3.  The  outside  of  a  pyramid  and  a  cone  is  called  the  lateral  or 
convex  surface,  the  area  of  which  may  be  found  by  multiplying  i  the  circum- 
ference or  girth  of  the  base  by  the  slant  hight ;  and,  when  the  entire  surface 
is  required,  to  the  product  adding  the  area  of  the  base.  Why  1 

8.  The  slant  hight  of  a  pyramid  is  11  inches,  and  the  base  is  4  inches 
square  ;  what  is  the  entire  surface  ? 

9.  What  is  the  area  of  a  triangular  pyramid,  each  side  oi  the  base 
measuring  30  feet,  and  the  slant  hight  42  feet  ? 

Ans.  2279<7114-{-sq.  ft. 

10.  What  is  the  lateral  surface  of  a  cone,  the  slant  hight  being  38 
inches,  and  the  circumference  of  the  base  40  inches  ? 

Ans.  5  sq.  ft.  40  sq.  in. 

11.  The  solidity  of  a  cone  is  214<S7235  cubic  feet,  and  the  altitude  16<75 
feet ;  what  is  the  diameter  of  the  base  ? 

NOTE  4.  This  example  involves  the  principles  of  the  rule,  but  they  must 
be  applied  in  a  reverse  order. 

12.  The  cubic  contents  of  a  square  pyramid  are  3000  cubic  inches,  and 
the  altitude  is  40  inches ;  what  is  the  length  of  one  side  of  the  base  ? 

13.  The  area  of  the  base  of  a  hexagonal  pyramid  is  259*8076  square 
yards,  and  the  soliditv  1299*038  cubic  yards ;  what  is  the  altitude  ? 

Ans.  45  ft. 

IT  81.  To  find  the  hight  of  a  pyramid  or  cone,  of  which 
a  given  frustrum  is  a  part. 

RULE. 

I.  For  the  pyramid  ;  —  Say,  the  difference  between  one  side 
of  the  top  and  one  side  of  the  base  :  one  side  of  the  base  : :  the 
altitude  of  the  frustrum  :  the  altitude  of  the  pyramid.     Or, 

The  difference  between  the  perimeter  of  the  top  and  the 
perimeter  of  the  base  :  the  perimeter  of  the  base  : :  the  altitude 
of  the  frustrum  :  the  altitude  of  the  pyramid. 

II.  For  the  cone  ; —  Say,  the  difference  between  the  diame- 
ters (or  the  radii)  of  the  top  and  base  :  the  diameter  (or  the 

IT  81.    Topic.    Analysis.    Rule. 


1182. 


MENSURATION  OF  SOLIDS.  83 


radius)  of  the  base  : :  the  altitude  of  the  frustrum  :  the  altitude 
of  the  cone.     Or, 

The  difference  between  the  peripheries  ot  the  top  and  base  . 
the  periphery  of  the  base  : :  the  altitude  of  the  frustrum  I  the 
altitude  of  the  cone. 

NOTE.    For  analysis  of  principles,  see  IT  44. 

EXAMPLES  FOR  PRACTICE. 

1  The  base  of  the  frustrum  of  a  pyramid  is  8  feet  square,  the  top  3 
feet  square,  and  the  altitude  15  feet  j  what  was  the  bight  of  the  pyramid  < 

2  The  base  of  the  frustrum  of  a  hexagonal  pyramid  is  22  inches  on 
each  side,  the  top  9  inches  on  each  side,  and  the  altitude  5  feet  ;  what 
was  the  altitude  of  the  pyramid  ?  . 

3  The  perimeter  of  the  base  of  a  decagonal  frustrum  is  8  feet 
inches,  the  perimeter  of  the  top  2  feet  1  inch,  and  the  altitude  1 
what  was  the  altitude  of  the  pyramid  ? 

4  The  diameter  of  the  base  of  the  frustrum  of  a  cone  is  5  feet,  the 
diameter  of  the  top  4  feet,  and  the  altitude  20  feet ;  what  was  the  altitude 

5  The  radius  of  the  base  of  the  frustrum  of  a  cone  is  17  inches,  the 
radius  of  the  top  14  inches,  and  the  altitude  5  inches  ;  what  was  the  z 
tude  of  the  cone  ?  „ 

6  The  circumference  of  the  base  of  the  frustrum  of  a  cone  is  4/  ieet, 
the  circumference  of  the  top  41  feet,  and  the  altitude  28§  feet ;  what  was 
the  altitude  of  the  cone  ?  A™.  224  it.  6}  in. 

IF  82.     To  find  the  solidity  of  the  frustrum  of  a  pyramid  or 
cone. 

RULE. 

I.  Find  the  hight  of  the  pyramid  or  cone  of  which  the 
given  frustrum  is  a  part,  by  If  81. 

II.  Find  the  cubic  contents  of  the  pyramid  or  cone,  and  also 
of  the  segment,  by  IT  80. 

III.  Subtract  the  cubic -contents  of  the  segment  from  the 
cubic  contents  of  the  entire  pyramid  ;  the  remainder  will  be  the 
cubic  contents  of  the  frustrum. 

NOTE  1.    The  pupil  will  readily  comprehend  the  reasons  for  each  step  in 
the  rule. 

EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  solidity  of  the  frustrum  of  a  square  pyramid,  one  side 
of  the  greater  end  being  18  inches,  one  side  of  the  smaller  end  15  inches, 
and  the  altitude  5  feet  ?  Ans.  9f  f  cu.  ft. 

2.  What  is  the  solidity  of  the  frustrum  of  a  hexagonal  pyramid,  the 
side  of  the  greater  end  beins:  3  feet,  that  of  the  smaller  end  2  feet,  and 
the  altitude  12  feet  ?     (See  f  54.) 

IT  82.    Topic.    Analysis.    Rule.    Note  2. 


84 


MENSURATION  OF  SOLIDS. 


183. 


3.  What  is  the  solidity  of  the  frustrum  of  a  cone,  the  diameter  of  the 
greater  end  being  4  feet,  that  of  the  smaller  end  2A  feet,  and  the  altitude 

/e<?Ji  mches  ?  Ans-  94-'9843-f  cu.  ft. 

4.  What  is  the  solidity  of  the  frustrum  of  a  cone,  the  circumference 
of  the  greater  end  being  83  inches,  that  of  the  smaller  end  54  inches,  and 
the  altitude  12  feet  ? 

5.  A  man  has  a  vessel  in  the  form  of  the  frustrum  of  a  square  pyra- 
mid ;  the  lower  end  is  30  inches  square,  the  upper  end  20  inches  square, 
and  the  altitude  4  feet  j  how  many  dry  gallons  will  it  contain  ? 

,,  Am.  113J  dry  gal. 

b.  Ihe  diameter  of  the  top  of  a  tub,  in  the  form  of  an  inverted  frustrum 
ol  a  cone,  is  40  inches,  the  diameter  of  the  bottom  30  inches,  and  the  alti- 
tude 5  feet  j  what  are  the  contents  in  standard  gallons  ?  Ans.  251'6  gal. 

NOTE  2.  The  lateral  surface  of  the  frustrum  of  a  pyramid  or  a  cone  may 
be  found,  by  multiplying  i  the  sum  of  the  girths  of  the  two  ends  by  the  slant 
night,  and  when  the  entire  surface  is  required,  to  the  product  adding  the 
areas  of  the  two  ends.  Why  ? 

7.  What  is  the  area  of  the  frustrum  of  a  pyramid  whose  slant  hight  is 
o  fee*,  the  base  4  feet  square,  and  the  top  2  feet  3  inches  square  ? 

Ans.  121  sq.  ft.  9  sq.  in. 

8.  What  is  the  area  of  the  frustrum  of  a  triangular  pyramid,  whose 
slant  hight  is  16  inches,  each  side  of  the  base  3  feet,  and  each  side  of  the 
top  1£  feet  ? 

9.  What  is  the  convex  surface  of  the  frustrum  of  a  cone,  whose  slant 
hight  is  18  inches,  the  circumference  of  the  base  38  inches,  and  of  the 
top  28  mches  ?  Ans,  4  sq>  ft>  18  sq.  in> 

IT  83.  To  find  the  superficial  and  the  cubic  contents  of  the 
regular  solids. 

ANALYSIS.  Each  of  the  regular  solids  may  be  divided  into  as  many 
equal  pyramids  as  the  solid  has  faces,  the  base  of  each  pyramid  being  a 
face  of  the  solid,  and  the  altitude  the  perpendicular  distance  from  the 
centre  of  one  face  to  the  centre  of  the  solid. 

Since  it  is  somewhat  difficult  to  find  the  altitude  of  the  pyramids  of 
which  each  regular  solid  is  composed,  the  following  table  has  been  pre- 
pared, by  the  aid  of  which  the  superficies,  and  the  solidity  of  any  regular 
solid,  may  readily  be  found,  by  having  one  side  and  the  number  of  sides 
given. 


Length  ol 
one  side. 

Number  of  sides. 

Names  of  Solids. 

Superficies. 

Solidity. 

1 

4 

Tetraedron, 

1  '73-2051 

'117851 

1 

6 

Hexaedron, 

6<000000 

I'OOOOOO 

1 

8 

Octaedron, 

3<464102 

'471404 

1 

12 

Dodecaedron, 

20-645729 

7-663119 

1 

20 

Tcosaedron, 

8<660254 

2-181695 

It  has  been  shown,  Tf^f  52  and  77,  that  the  areas  of  similar  figures  are 
to  each  other  as  the  squares  of  their  similar  dimensions.     It  is  also  true 


IT  83.    Topic.     Analysis.     Reasons  for  preparing  a  table.     Its  construc- 
tion.   Similar  solids.    Rule  for  use  of  table. 


IT  84.  MENSURATION  OF  SOLIDS.  85 

that  the  solidities  of  similar  bodies  are  to  each  other  as  the  cubes  of  their 
similar  dimensions.*    Hence, 

To  find  the  superficies  or  the  solidity  of  any  regular  solid,  ly 
the  table. 

*   RULE. 

I.  For  the  superficies  ;  —  Multiply  the  square  of  one  side  by 
the  tabular  number  of  the  superficies  of  a  similar  solid. 

II.  For  the  solidity  ;  —  Multiply  the  cube  of  one  side  by  the 
tabular  number  of  the  solidity  of  a  similar  solid. 

EXAMPLES  FOR  PRACTICE. 

1.  The  side  of  a tetraedron measures  7  inches  5  what  are  its  superficies 
and  solidity  ? 

2.  The  side  of  an  octaedron  measures  4  inches ;  what  are  its  superficies 
and  solidity  ? 

3.  What  are  the  superficies  and  the  solidity  of  a  dodecaedron,  one  side 
of  which  measures  4£  feet  ? 

4.  What  are  the  superficies  and  the  solidity  of  an  icosaedron,  one  side 
of  which  measures  3  inches  ? 

IF  84.     To  find  the  cubic  contents  of  any  irregular  solid. 

RULE. 

I.  Place  the  solid  in  a  tub,  cylinder,  cubical  box,  or  any 
other  vessel  whose  contents  can  be  ascertained,  and  then  fill 
the  vessel  with  water. 

II.  From  the  cubic  contents  of  the  vessel  subtract  the  cubic 
contents  of  the  water  put  in  to  fill  the  vessel ;  the  remainder 
will  be  the  cubic  Contents  required. 

NOTE.  Any  vessel  may  be  filled  with  water,  and  the  body  whose  contents 
are  required  may  then  be  immersed  in  the  water  ;  the  quantity  of  water  Avhich 
the  body  displaces,  or  which  runs  over  the  sides  of  the  vessel,  will  be  equal  in 
bulk  or  cubic  contents  to  the  figure  immersed. 

EXAMPLES  FOR  PRACTICE. 

1.  A  blacksmith's  anvil  was  put  into  a  tub,  the  capacity  of  which  was 
8£  wine  gallons,  and  the  tub  was  afterwards  filled  with  6  gal.  3  qts.  1  pt. 
of  water  ;   what  was  the  solidity  of  the  anvil?  Ans.  317|  cu.  in. 

2.  A  chain  was  put  into  a  cubical  box  whose  inside  measured  8  inches, 
and  the  box  was  afterwards  filled  with  3|  quarts  of  water ;  what  were 
the  cubic  contents  of  the  chain  ?  Ans.  395T7F  cu.  in. 

3.  A  pig  of  iron  was  put  into  a  cylinder  3  feet  long,  and  7  inches  in 

IF  84.    Topic.     Analysis.     Rule.     Method  explained  in  note. 

*  It  may  be  well  to  test  the  pupil's  comprehension  of  this  truth,  by  giving 
him  a  few  practical  examples  in  Similar  Solids.     See  TTTT  52  and  77. 
8 


86  MENSURATION  OF  SOLIDS.  H  85,  86. 

diameter,  and  the  cylinder  was  then  filled  with  4  ^^Jg^.  £* 
were  the  cubic  contents  of  the  iron? 

If  85.     To  find  the  area  of  a  sphere  or  globe, 
It  is  demonstrated  in  Geometry,  Aat  the  area  of  a  sphere  or 
globe  is  4  times  as  great  as  the  are'a  of  a  circle  of  the  same 
diameter.     Hence, 

To  find  the  area  of  a  sphere. 

RULE. 

Multiply  4  times  the  square  of  the  diameter  by  '7854.     Or, 
Multiply  the  whole  circumference  by  the  whole  diameter. 

NOTE.     A  knowledge  of  Geometry  is  necessary  to  a  full  understanding  of 
the  principles  of  this  rule. 

EXAMPLES  FOR  PRACTICE. 

1.  How  many  square  inches  on  the  surface  of  a  globe  15  inches  in 
many  square  feet  on  the  surface  of  a  sphere   4  feet  in 

iameter  of  the  earth  is  7911  miles  ;  what  is  the  area  rejecting 
fractions  of  a  mile  in  the  circumference? 

IF  86.     To  find  the  solidity  of  a  sphere. 

ANALYSIS.  Any  sphere  may  be  supposed  to 
be  divided  into  an  infinite  number  of  pyramids, 
whose  vertices  all  meet  in  the  centre  of  the 
sphere,  and  the  areas  of  whose  bases  form  the 
area  of  the  sphere.  Since  the  solidity  of  any 
pyramid  is  equal  to  the  area  of  its  base  multiplied 
by  I  of  its  altitude,  (H  80,)  the  solidity  of  all  the 
pyramids  into  which  any  sphere  may  be  supposed 
to  be  divided,  is  equal  to  the  areas  ot  all  their  ,  . 

ba^es  (winch  is  the  surface  of  the  sphere)  multiplied  by  \  of  the  alti 
tude  of  one  of  them,  or  £  of  the  diameter  of  the  sphere.     1! 

To  find  the  solidity  of  a  sphere. 

RULE. 

Multiply  the  area  of  the  sphere  by  £  of  the  diameter. 


IT  85.    Topic.    Analysis.     Rule.    Note. 
IT  86.    Topic.    Analysis.    Rule.    Note  1 


MENSURATION  OF  SOLIDS.  87 

to j  cube -of the  diameter  of  the  sphere  by  '7354,  and  taking  \  of  the  product. 
EXAMPLES  FOR  PRACTICE. 

1.  The  diameter  of  a  sphere  is  18  inches  ;  what  is  its  solidity  ? 
2    What  is  the  solidity  of  an  ivory  ball  2  inches  m  diametei  ? 
J  What  is  the  solidity  of  a  wicket  ball  18*  inc,he1circU 

4.  The  diameter  of  the  earth  is  7911  miles  ;  w^ 

ing  fractions  of  amile  in  the  circumference  ?    Ans.  2o , 

5.  What  is  the  solidity  of  a  ball  that  can  just  be  put  into  a  cylindrical 
cup  5  inches  in  diameter,  and  5  inches  deep  ? 

6.  What  is  the  solidity  of  a  hemisphere  12  ""^jJ^gJEj.  cu.  in. 

7.  The  solidity  of  a  sphere  is  65<45  cu.  in. ;  what  is  its  diameter? 
NOTE  2.    This  example  may  be  performed  by  the  principles  given  in  Note 

1  ;  but  they  must  be  applied  in  a  reverse  order. 


GAUGING. 

IT  87.     Gauging  is   measuring  the  capacity   of  barrels, 
casks,  hogsheads,  &c. 

A™  The  mean  diameter  of  a  barrel,  cask,  &c.,  may  be  found, 

by  adding  to  the  heTd  diameter  |,  or,  if  the  staves  be  but  little  jcorvmft 

will  give  the  cubic  contents  in  cubic  inches.    . 

To  gauge  or  measure  a  cask. 

RULE. 

I    For  the  capacity  in  cubic  inches;—  Multiply  the  area  of 
the  mean  diameter  in  inches  by  the  length  in  inch. 

II.  For  the  capacity  in  standard  or  wine  gallons,  —  Ui 
the  capacity  in  cubic  inches  by  231. 

Ill    For  the  capacity  in  bushels;  -Divide  the  capacity   m 
cubic  inches  by  2150'4.     IT  24. 


EXAMPLES  FOR  PRACTICE. 

1.  The  head  diameter  of  a  cask  is  22   inches,   the  bung 

U  87.    Topic.    Gauging.    Analysis.    Rule.    Note. 


88  MENSURATION  OF  SOLIDS.  f  88, 89. 

28  inches,  and  the  length  31  inches ;  how  many  standard  or  wine  gallons 
will  it  contain  ?    How  many  bushels  ?    How  many  beer  gallons  ? 

Ans.  to  first  two.   71'2504  wine  gal. ;  7'6538-f-  bush. 
2.  The  head  diameter  of  a  cask  is  30  inches,  the  bung  diameter  35 
inches,  and  the  length  40  inches ;  what  is  its  capacity  in  standard  li- 
quid gallons  ?  —  in  bushels  ? 


TIMBER  MEASURE. 

IT  88.  Square  or  hewn  timber  is  sometimes  bought  and 
sold  by  the  cubic  foot,  and  is  sometimes  reduced  to  standard 
board  measure. 

To  find  the  number  of  cubic  feet  in  any  stick  of  hewn  timber 
which  does  not  taper. 

For  rule  and  principles,  consult  f  79. 

To  find  the  number  of  feet,  board  measure,  in  any  stick  of 
hewn  timber  which  does  not  taper, 

For  rule  and  principles,  see  f  IT  79  and  37. 

EXAMPLES  FOR  PRACTICE. 

1.  How  many  cubic  feet  in  a  stick  of  timber  50  feet  long,  and  7  by  10 
inches  ? 

2.  How  many  cubic  feet  in  a  stick  of  timber  40  feet  long,  and  22  by 
27  inches  ? 

3.  How  many  feet,  board  measure,  in  a  stick  of  timber  60  feet  long, 
and  8  by  14  inches  ? 

4.  How  many  feet,  board  measure,  in  a  stick  of  timber  35  feet  long, 
15  inches  wide,  and  12  inches  thick  ? 

IT  89.  To  find  the  contents  of  a  four -sided  stick  of  timber, 
which  tapers  upon  two  opposite  sides  only. 

ANALYSIS.  If  a  stick  of  timber  tapering  upon  two  opposite  sides  only, 
be  sawn  into  boards  in  a  line  perpendicular  to  the  tapering  sides,  it  will 
make  a  certain  number  of  boards  of  uniform  length,  and  all  tapering 
alike.  Hence, 

To  find  the  contents  of  a  four-sided  stick  of  timber  which 
tapers  upon  the  opposite  sides  only. 

RULE. 

I.  Divide  the  sum  of  the  widths  of  the  two  ends  of  either 

IT  88.    Measure  by  which  square  timber  is  bought  and  sold.    First  rule. 
Analysis.    Second  rule.     Analysis. 
U89.    Topic.    Analysis.    Rule. 


IT  90,  91.  MENSURATION  OF  SOLIDS. 

tapering  side  by  2  ;  the  quotient  will  be  the  mean  width  of  the 
tapering  side. 

II.  Proceed  in  all  other  respects  as  directed  in 

EXAMPLES  FOR  PRACTICE. 

1.  A  stick  of  timber  is  24  feet  long,  15  inches  thick,  ££? 
one  end   and  8  at  the  other;  what  are  its  cubic  contents? 

^JK3«roV**»  ^  inches  wide,  15  niches  .hick  at 
one  end,  and  10  at  the  other  ;  what  are  its  cubic  contents  ?     What  i 
contents  in  board  measure  ? 

IT  OO.  To  find  the  contents  of  a  stick  of  timber  which  tapers 
uniformly  upon  all  sides. 

ANALYSIS.  A  stick  of  timber  tapering  uniformly  upon  all  sides,  is 
either  a  pyramid  or  the  frustrum  of  a  pyramid  and  consequently  mu,t 
be  measured  by  the  same  principles.  See  f  f  80  and  tw. 

EXAMPLES  FOR  PRACTICE. 

1  A  stick  of  hewn  timber  is  31<5  feet  long,  18  inches  square  at  one 
end,  and  8  inches  square  at  the  other  ;  what  are  its  cut  ? 

>'.  A  stick  of  timber  20  feet  long  is  12  inches  square  at  one  end,  K 
inches  square  at  the  other,  and  20  inches  square  in  the  ,  midd  e  ; 
its  contents  in  board  measure  ?  Ans.  4/4  sq.  ft.  104  sq.  in. 

IF  91.  To  find  the  number  of  cubic  feet  of  timber  any  log 
will  make  when  hewn  square. 

RULE. 

Find  the  area  of  the  inscribed  square  of  the  smaller  end  of 
the  log,  and  multiply  this  area  by  the  length. 


founding 

mog  for^y  purpose,  it  is  considered  as  a  cylinder 
of  the  same  diameter  as  the  smaller  end  of  the  log.  Hence,  the  diamete  rol 
the  smaller  end  of  the  log  must  always  he  taken  ;  and,  if  the  end  be  elliptical, 
the  shorter  diameter  must  he  taken,  not  the  longer  one. 

EXAMPLES  FOR  PRACTICE. 

1.  A  log  is  28  feet  long,  and  20  inches  in  diameter  ;  how  many  feet  of 
timber  will  it  make  when  hewn  square  ? 

IT  90.   Topic.    Analysis.    Principles  referred  to.    Rules  for  the  operations. 
U  91.    Topic.    Analysis.    Rule.    Note  2.    Note  3. 


9Q  MENSURATION  OF   SOLIDS.  f  92. 

2.  A  log  20  feet  long,  and  10  inches  in  diameter,  was  hewn  square ; 
how  many  cubic  feet  were  cut  away  ?  Ans.  3  cu.  ft.  1665<6  cu.  in. 

3.  A  stick  of  timber  12  feet  long,  and  14  inches  in  diameter,  was 
hewn  into  a  hexagonal  form:  how  many  cubic  feet  did  the  hexagon 
contain?  Ans.  10'60881-j-  cu.  ft. 

IT  92.  To  find  the  number  of  feet  of  boards  that  can  be 
sawn  from  any  log  of  a  given  diameter. 

Logs  for  sawing  are  measured  in  three  different  ways,  which 
we  will  designate  as  1st,  2d,  and  3d  methods. 

First  Method. 

ANALYSIS.    The  allowances  to  be  made  in  this  method  are. 

1st.  For  slabs.  This  is  an  allowance  of  2  inches  on  each  side,  or  4 
inches  of  the  diameter  of  any  log  not  exceeding  2  feet  in  diameter  ;  and 
3  inches  on  each  side,  or  6  inches  of  the  diameter  of  any  log  more  than  2 
feet  in  diameter. 

2nd.  For  saw  space.  This  is  an  allowance  of  £  of  an  inch  for  each 
time  the  saw  passes  through  the  log.  In  sawing  boards  of  the  standard 
thickness,  the  saw  cuts  away  just  i  of  the  log  after  the  slabs  are  re- 
moved. 

3d.  For  mane.  This  is  an  allowance  of  1  board  for  any  log  not  ex- 
ceeding 2  feet  in  diameter ;  and  of  2  boards  for  any  log  more  than  2  feet 
in  diameter.  This  allowance  is  made  to  cover  the  loss  that  would 
otherwise  arise,from  estimating  the  wane-edged  boards  the  same  as  those 
that  are  square-edged. 

EXAMPLE.  How  many  feet  of  boards  can  be  cut  from  a  log  12  feet 
long,  and  2  feet  =  24  inches  in  diameter  ? 

SOLUTION.  24  in.  (diameter)  —  4  in.  (slabs)  =  20  in.  for  sawing.  20 
in.  —  4  in.  (  £  of  20  in.  for  saw  space)  =  16  in.  thickness  of  all  the  boards, 
which  —  1  (for  wane)  =  15  boards  that  measure.  We  now  have  the 
log  reduced  to  a  stock  of  15  boards,  each  12  feet  long,  and  20  inches 
wide.  20  in.  (width  of  1  board)  X  15  (no.  of  boards)  =  300  in.  =  25  ft., 
the  whole  width  of  the  boards.  Then,  25  ft.  (width)  X  12  ft-  Oengtn)  = 
300  sq.  ft.,  the  Ans.  Hence, 

For  the  first  method. 

RULE. 

I.  Make  the  customary  allowances  for  slabs,  saw  space,  and 
wane  ;  the  remainder  is  the  number  of  standard  boards  that  can 

IT  92.  Topic.  Methods  of  measuring  logs  for  sawing.  Analysis  of  first 
method.  1st  allowance.  2d.  3d.  Solution  of  example.  Rule. 


H  93,94.  MENSURATION  OF  SOLIDS.  91 

be  sawn  from  the  log,  the  width  of  each  being  the  diameter  of 
the  log  minus  the  allowance  for  slabs. 

II.  Find  the  contents  of  the  stock  of  boards  to  which  the 
log  is  reduced,  as  directed  in  H  57. 

IT 93.    Second  Method. 

ANALYSIS.  In  this  method  no  deductions  or  allowances  are  made,  but 
the  log  is  reduced  to  a  stick  of  square  timber,  by  H"  91,  and  then  to  board 
measure.  It  is  estimated  that  in  reducing  the  log  to  a  stick  of  square 
timber,  the  siding  or  wane-edged  boards  will  equal  in  quantity  the  loss  by 
saw  space.  In  this  estimate  there  is  a  little  advantage  m  lavor  ot 
buyer  •  but  when  we  consider  that  some  logs  are  crooked,  some  rotten, 
some  hollow,  &c.3  it  seems  but  just  to  have  the  advantage  in  his  favor. 
Hence, 

For  the  second  method. 

RULE. 

Find  the  number  of  feet,  board  measure,  that  the  log  will 
make  when  hewn  square. 

NOTE.     This  method  is  the  one  generally  adopted  in  the  lumbering  districts 
of  New  York  and  Pennsylvania.     Some  of  the  most  accurate  business  men  m 
the  lumber  trade,  after  trying  various  methods,  say,  that  this  method  i 
nearly  correct  for  all  logs  over  13  inches  in  diameter,  as  any  other  method 
that  has  been  presented. 

IF  94.    Third  Method. 

ANALYSIS.  In  this  method  a  slab  of  1  inch  in  thickness  is  first  taken 
off  from  one  side,  and  the  log  is  then  sawn  up  into  wane-edged  boards. 
The  slab  upon  the  opposite  side  to  the  one  first  taken  off,  is  never  allowed 
to  be  less  than  |  of  an  inch,  nor  over  2  inches,  in  thickness.  From  an 
actual  measurement  of  stocks  of  boards  sawn  from  logs  varying  from  7 
to  36  inches  in  diameter,  by  this  method,  it  has  been  found  that 

1st.  The  width  of  the  second  board  is  the  average  width  of  all  the 
board's  sawn  from  any  log  from  7  to  12  inches  in  diameter. 

2nd.  The  width  of  the  third  board  is  the  average  width  of  all  the 
boards  sawn  from  any  log  from  12  to  24  inches  in  diameter.  And 

3rd.  The  width  of  the  fourth  board  is  the  average  width  of  all  the 
boards  sawn  from  any  log  from  24  to  36  inches  in  diameter.  Hence, 

IT  93.    Analysis  of  second  method.    Rule.     Note. 

IT  94.  Analysis  of  third  method.  Stocks  of  boards  sawn  from  logs  from 
7  to  12  inches  in  diameter,  by  this  method.  From  12  to  24  inches  in  diame- 
ter. From  24  to  36  inches.  Rule.  Note  1.  Note  2. 


92  MENSURATION  OF  SOLIDS.  IT  95. 

For  the  third  method. 

RULE. 

I.  Multiply  the  width  of  the  average  board,  in  inches,  by 
the  number  of  boards. 

II.  Multiply  this  product  by  the  length  of  the  log  or  stock 
of  boards  in  feet,  and  divide  the  product  by  12. 

NOTE  1.     Consult  H  58. 

_  NOTE  2.    The  manner  of  sawing  logs  described  in  this  IT  is  in  quite  exten- 
sive practice  in  New  England. 

EXAMPLES  FOR  PRACTICE. 

1.  The  diameter  of  a  log  is  31  inches,  and  the  length  13  feet  j  how 
many  feet  of  boards  will  it  make,  estimating  by  the  first  method  ?    By 
the  second  ?  Arts.  By  1st  method,  487£  sq.  ft, ;  by  2d,  520j-f  sq.  ft. 

2.  The  diameter  of  a  log  is  40  inches,  and  the  length  10  feet ;  how 
many  feet  of  boards  will  it  make,  estimating  by  the  first  method?    By 
the  second?  Answers,  in  order.     708£  sq.  ft. ;  666|  sq.  ft. 

3.  How  many  feet  of  boards  may  be  sawn  from  a  log  19  inches  in  di- 
ameter, and  16  feet  long,  estimating  by  the  first  method  ?     By  the  2d  ? 

Answers,  in  order.     220  sq.  ft. ;  240|  sq.  ft. 

IT  95.     To  find  how  many  bushels  of  grain  may  be  put 
into  a  bin  of  a  given  size. 

RULE. 

Divide  the  cubic  contents  of  the  bin  in  inches  by  21o0'4. 

'  NOTE  1.     If  the  pupil  does  not  readily  recognize  the  principles  involved 
in  the  rule,  he  should  review  1T1T  24  and  79. 

NOTE  2.  2150{4  cu.  in.  (  =  1  bush.)  — 430^8  cu.  in.  (  =  A  of  a  bush.)  = 
1720!32  cu.  in.,  or  7'68  cu.  in.  less  than  one  cubic  foot.  Hence,  if  the  cubic 
feet  in  any  bin  be  diminished  -1-,  the  remainder  will  be  the  number  of  bushels 
which  the  bin  will  contain.  This  result,  although  not  strictly  correct,  is  suffi- 
ciently accurate  for  all  practical  purposes. 

EXAMPLES  FOR    PRACTICE. 

1.  I  have  a  bin  5  feet  long,  4  feet  8  inches  wide,  and  2  feet  10  inches 
deep  ;  how  many  bushels  of  grain  will  it  hold  ? 

2.  A  farmer  has  a  bin  8  feet  long,  4  feet  wide,  and  5  feet  deep  j  how 
many  bushels  of  grain  will  it  hold  ? 

3;  A  bin  5  feet  long,  and  4  feet  8  inches  wide,  contains  53|  bushels  of 
grain  ;  what  is  its  depth  ? 

4.  A  farmer  wishes  to  construct  a  bin  8  feet  long,  and  5  feet  deep,  and 
he  will  have  it  hold  128  bushels  j  what  must  be  its  width  ? 

IT  95.    Topic.    Analysis.    Rule,    Note  2.    Note  3. 


1T  96,  97. 


MENSURATION  OF   SOLIDS. 


93 


NOTE  3.  1728  cu.  in.  (=  1  cu.  ft.)  -f-  432  cu.  in.  (=  i  of  a  cu.  ft  )  =2160 
cu.  in.,  or  9'G  cu.  in.  more  than  1  bushel.  Hence,  if  the  number  ot  bushels 
in  any  bin  be  increased  £,  the  sum  will  be  the  contents  in  cubic  feet.  This 
result,  although  not  strictly  correct,  is  sufficiently  accurate  for  all  practical 
purposes. 

IT  96.     TABLE, 

Showing  the  inside  dimensions  of  any  box  of  a  given  capacity, 
in  dry  measure. 


Length 


Width.       |        Depth.        | 


Capacity, 


4  inches. 

4  inches. 

4<2  inches. 

1  quart. 

7       « 

4       « 

4<8 

2  quarts  = 

£  gallon. 

8       " 

8       « 

4<8 

1  gallon  = 

^  peck. 

8*4    " 

8       " 

8 

1  peck. 

12       " 

8 

2  pecks  =  i 

bushel. 

16<8    « 

16       « 

8 

1  bushel. 

IT  97.     To  find  the  side  of  the  greatest  cube  that  can  be 
cut  from  any  sphere. 

ANALYSIS.  The  diameter  or  axis  of  the 
sphere  is  the  length  of  a  diagonal  running 
from  any  lower  corner  of  the  cube  to  the  di- 
agonal corner  at  the  top.  Let  us  first  see 
how  this  diagonal  of  any  cube  is  obtained. 
We  first  square  a  side  of  the  cube,  double 
it,  and  then  extract  the  square  root.  This 
gives  us  the  diagonal  of  one  face  of  the 
cube.  We  next  square  this  diagonal,  (which 
square  xvill  contain  twice  the  square  of  one 
side,)  add  to  it  the  square  of  one  side  of  the  cube,  (making  the  sum 
3  times  the  square  of  one  side  of  the  cube,)  and  then  extract  the  square 
root  of  this  sum.  Hence, 

To  find  the  side  of  the  greatest  cube  that  can  be  cut  from 
any  sphere. 

RULE. 

Divide  the  square  of  the  diameter  of  the  sphere  by  3,  and 
extract  the  square  root  of  the  quotient. 

EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  size  of  the  largest  cube  that  can  be  cut  from  a  sphere 
15  inches  in  diameter  ?  Ans.  8<66-j-  in. 

2.  I  have  a  globe  25  inches  in  diameter,  and  I  wish  to  cut  from  it  the 
largest  possible  cube ;  how  much  of  the  sphere  will  I  cut  away  ? 

'  Ans.  5794'98-4-cu.  in. 


IT  96. 
IT  97. 


Table. 
Topic. 


Analysis.    Rule. 


94  MENSURATION  OF  SOLIDS.  IF  98,  99. 

IT  98.     To  find  the  weight  of  a  lead  or  of  a  cast-iron  ball. 

ANALYSIS.  The  cubic  contents,  and  consequently  the  weights  of  sim- 
ilar bodies  of  the  same  substance,  are  to  each  other  as  the  cubes  of  their 
similar  dimensions.  (*f[^  52,  77,  and  83.)  A  leaden  ball  1  inch  in 
diameter  weighs  T3j-  of  a  pound  ;  and  a  cast-iron  ball  4  inches  in  diam- 
eter weighs  9  pounds.  Hence, 

To  find  the  weight  of  a  lead  or  of  a  cast-iron  laU. 

RULE. 

I.  For  a  leaden  ball ;  —  1 3  :  the  cube  of  the  given  diame- 
ter : ;  T3¥  lb.  :  the  weight  of  the  given  ball. 

II.  For  a  cast-iron  ball;  —  43  :  the  cube  of  the  given  di- 
ameter : :  9  Ibs.  :  the  weight  of  the  given  ball. 

EXAMPLES  FOR    PRACTICE. 

1.  What  is  the  weight  of  a  leaden  ball  5  inches  in  diameter  ? 

Ans.  26iJ  Ibs. 

2.  What  is  the  weight  of  a  cast-iron  ball  7  inches  in  diameter  ? 

Ans.  48£flbs. 

3.  A  case-iron  ball  weighs  22<5  pounds  ;  what  is  its  diameter  ? 

Ans.  5'42-j-in. 

PRACTICAL    EXAMPLES 

IN  THE  MENSURATION  OF  SOLIDS. 

5FO9.      1.   How  many  5  inch  cubes  are  equal  to  a  20  inch  cube? 

Ans.ftt. 

2.  A  stick  of  timber  is  30  feet  long,  10  inches  wide,  10  inches  thick  at 
one  end,  and  7  at  the  other;  what  are  its  cubic  contents  ? 

Ans.  17  cu.  ft.  102  cu.  in. 

3.  A  stick  of  timber  is  23  feet  long,  20  inches  thick.  14  inches  wide  at 
each  end,  and  20  in  the  middle ;  what  are  its  contents  in  board  measure  ? 

Ans.  651|  sq.  ft. 

4.  I  have  4  sticks  of  timber  of  the  same  length,  each  25  inches  wide, 
and  whose  thicknesses  are  21,  10,  6,  and  3'5  inches,  respectively.  I  wish 
a  square  stick  of  the  same  length,  and  containing  as  much  timber  as  the 
whole  4  ;  what  will  be  the  measure  of  one  side  ?  Ans.  31'81-f-  in. 

5.  How  many  cubic  feet  in  a  log  50  feet  long,  and  16  inches  in  diam- 
eter ?  Ans.  69-8 }  cu.  ft. 

6.  I  have  a  joist  8  inches  wide,  and  3  inches  thick.    I  want  a  stick  of 
timber  twice  as  wide,  and  containing  4%  times  as  much  timber ;  what 
must  be  its  thickness  ?  Ans.  6|  in. 

7.  A  man  dug  and  stoned  a  well  upon  the  following  terms,  viz. :   for 
the  first  15  feet  he  received  $1<00  a  foot,  for  the  next  10  feet  $1<50  a 
foot,  for  any  greater  depth  S2'00  a  foot,  and  for  every  foot  of  rock  through 

V  98.  Topic.  Analysis,  Applications  of  the  principles  contained  in 
HIT  52,  77,  and  83.  Rule,  for  a  leaden  ball;  —  for  a  cast-iron  ball. 


99  PRACTICAL  EXAMPLES. 


it  the  first  day  ;  on  which  day  did  ^    |s(  day;  gj  cu  £l. 

9.  What  is  ,he  area  of  a  cylinder  M  feet  long,  and  *£*$£%  £ 


is  8 
a 

3  feet  high,  and 


its  base  is  3  feet  ;  what  is  its  ^titude?  d]ameter  Qf 

14.  The  solidity  of  a  cone  is  8    &™c™  Ans.  15  ft. 


ThP  trpp  bein^  of  oak,  an  allowance  of  trr  of  the 


contents  of  the  tree  and  branches  ? 


PRACTICAL  EXAMPLES.  f  99. 


96 

of2ln'mhp!°iy SfJ3  feet,101lSVa^d  13  inches  in  diameter;  how  many  feet 
umber  ot  the  standard  thickness  will  it  make,  estimating  by  the  sec- 
ond method?  Tort  3         f 

Ans.   195rj-Q  sq.  ft. 

of  the  Sandajdnh'1(kgS  ^  inches  in  diameter.-  will  make  as  much  lumber 

Ans.  225. 
2  feet  in  diameter?1  be^  ^54^ 

inches  nnHth1PfS  d;SCha1!'ge  Waier  intO  a  third>  whose  diameter  is  13 
One  nf  ??  u*     Se    ?  are  sufficient  to  kgep  the  third  constantly  full. 

a  «?~«  r    Sma  f  P1PC?  dlscharges  twice  as  much  water  as  the  other  in 
a  given  time  ;  what  is  the  diameter  of  each  of  the  smaller  pipes  ? 

Ans.  One  is  7<5-}-  in. ;  the  other  10'61-f-  in. 

is  the  solidllvT  the  SUperficies  of  a  bomb  1G  inches  in  diameter  ?    What 

Ans.  to  last.     1  cu.  ft.  416<6656  cu.  in 

lian   re,W4ntlf?romy  P^  an  ir°n  roller'  which  sha11  be  20  inches  in 
ter,  4  leet  2  inches  long,  and  the  metal  I£  inches  thic^      «n«-:-- 

i  to  weigh  4^  ounces,  what  will 


vht-          '  church-'  in  London'  is6  ' 

what  did  the  gilding  cost,  at  5  cents  per  square  inch? 

Ans.  $814*30. 
64.  A  cast-iron  ball  weighs  72  pounds  ;  what  is  its  diameter  ? 

?3;v,TheuTalls  °f  a  house  are  each  27'5  feet  long,  and  26  feet'hi-h' 
and  the  gables  are  10  feet  high.     10  feet  high  of  the^waJls  are  2k  bricks 

h  '  L   fl       m°J5i2  bncks  thick'  6  feel  ^  bricks  lhick>  and  ^  ^bles 
brick  thick.     What  will  the  materials  and  bricklaying  come  u>    a? 
per  square  rod  ?    (See  U  17,  Note  2  ;  and  If  18,  Note  1.) 


. 

Pp  M        5°  feet  long'  34  feet  wide«  20  feet  hi§h  to  the 

are  2  Vp^ti    i  g\bleS  nSe  9,  feet  ab°Ve  the  eaves-     9  feet  of  ^e  walls 
et  thick  the  remainder  to  the  eaves  1£  feet  thick,  and  the  gables 
oo  .thick.     How  many  cubic  feet  in  the  walls?     What  did  the  build- 


t  >  - 

35  A  man  built  a  brick  house  36  feet  long,  28  feet  wide,  and  18  feet 
high  to  he  eaves  ;  and  the  walls  were  1  foot&thick.  How  many  bricks 
did  it  take  lor  the  walls.,  allowing  each  brick  to  be  8  inches  long,  4  inches 
wide,  and  2  inches  thick-  and  making  no  deductions  for  windows,  doors 
°r^01  ' 


Ans.  60264' 

Jb.  How  many  bricks  would  it  take  to  build  the  above  house,  allowino- 

^  of  an  inch  upon  three  sides  of  each  brick,  for  the  thickness  of  the* 

::'        '  '  Ans.  48889-f  bricks. 

o7    A  laborer  dug  and  stoned  a  cellar  21  feet  long,  16  feet  wide,  and 

iff'  Up°n     e  followi»g  terms,  viz.  :  he  received  S'18|  per  cubic 
yard  for  digging,  $«50  per  perch  for  ]aying  what  stone  he  fo|^      ™£ 

gmi;  ami  .^24  per  perch  for  the  remainder.  In  digging  the  cellar  he 
found  11^  perch  of  stone.  The  walls  of  the  cellar  being  2  feet  thick 
how  much  did  he  receive  for  his  labor  ?  (See  f  18.)  AM.  $41<012+: 


• 


IT  99.  PRACTICAL  EXAMPLES.  97 

38.  The  head  diameter  of  a  cask  is  16£  inches,  the  bung  diameter  21<6 
inches,  and  the  length  26  inches  ;  what  is  its  capacity  in  standard  gal- 
lons? Ans.  35<007284  gal. 

39.  I  have  a  granary  17  feet  6  inches  long,  8  feet  9  inches  wide,  and 
5  feet  3  inches  deep.     I  wish  to  construct  another  whose  dimensions 
shall  be  in  a  similar  proportion,  and  which  shall  hold  8  times  as  much  ; 
what  will  be  the  dimensions  of  the  new  one,  and  what  its  capacity  in 
bushels  ? 

Ans.   Dimensions,  35  ft.  long,  17  ft.  6  in.  wide,  10  ft.  6  in.  deep ; 
capacity,  5167<96£  bush. 

40.  A  church  spire  is  to  be  built  in  the  form  of  a  hexagonal  pyramid, 
one  side  of  the  base  being  10  feet,  and  the  altitude  80  feet.     Within  the 
spire  is  to  be  a  hollow  cone  15  feet  in  diameter  at  the  base,  and  so  run- 
ning up  as  to  leave  the  walls  of  the  spire  as  thick  at  the  top  of  the  cone 
as  at  the  bottom.    How  many  cubic  feet  will  the  spire  contain  ? 

Ans.  3393<902§  cu.  ft. 

9 


MECHANICAL  POWERS. 

If  1OO.  The  Mechanical  Pmvers  are  instruments  or  simple 
machines,  employed  to  facilitate  the  moving  of  weights  or  the 
overcoming  of  resistance.  They  are  six  in  number;  viz., 
the  Lever,  the  Wheel  and  Axle,  the  Pulley,  the  Inclined  Pla?ie, 
the  Wedge,  and  the  Screw. 

In  mechanical  powers  and  in  machinery,  the  thing  to  be 
moved,  or  the  resistance  to  be  overcome,  is  called  the  Weight; 
and  the  force  which  is  applied  to  effect  the  object,  is  called  the 
Power. 


The  Lever. 

f  1O1.  The  Lever  is  an  inflexible  bar  or  rod  supported 
at,  and  movable  about  a  point. 

The  Fulcrum  or  Prop  is  the  point  upon  which  a  lever  is 
sustained,  and  about  which  it  moves.  The  distances  from  the 
fulcrum  to  the  points  of  the  lever  at  which  the  weight  and 
power  act,  are  called  the  Arms  of  the  lever. 

Levers  are  commonly  divided  into  three  kinds,  according  to 
the  relative  positions  of  the  power,  the  weight,  and  the  fulcrum. 

In  a  lever  of  the       A                                  B  c 

first  kind  the  ful-        j — -T- "     " 


crum  F  is  be- 
tween  the  power 
P  and  the  weight 
W.  AB  is  the 
long  arm,  and  BC 
the  short  arm. 

IT  100.    Topic.    Mechanical  powers.    Their  numbe;  and  names.    Weight. 

Power. 
IT  101.    Topic.    Lever.    Fulcrum  or  prop.    Arms  of  the  lever, 


IT  102. 


MECHANICAL  POWERS. 


j.n  a  lever  of  the 
second  kind  the 
weight  W  is  be- 
tween the  power 
P  and  the  fulcrum 
F.  AC  is  the  long 
arm,  and  BC  the 
short  arm. 


In  a  lever  of  the 
third  kind  the  pow- 
er P  is  between 
the  weight  W  and 
the  fulcrum  F. 
BC  is  the  long 
arm,  and  AC  the 
short  arm. 


In  making  estimates  of  the  force  of  the  lever,  four  things  are 
always  considered;    viz.,  the  long  arm   the  shor 
power,  and  the  weight;  any  three  of  which  being  given,  tt 
other  may  be  found. 

V  102  It  is  a  fixed  principle  in  mechanics,  that  a  lever 
JaMt  at  rest,  when  the  ratio  of  the  weight  to  the  poiocru 
epud  to  the  ratio  of  the  long  arm  to  the  short  arm  In  other 
words  a  lever  will  remain  at  rest,  when  the  weight  X  fy  its 
Stance  from  the  fulcrum  =  the  power  X  *»  «,  distance  from 


W  represent  the  weight,  P  the  power,  LA  the 
arm,  and  SA  the  short  arm,  the  several  processes  may  be 


the  weight. 

r_. 

SA 

of  lever.     1st  kind.    2d  kind.    3d  kind. 
ered  in  making  estimates  upon  the  lever. 


Number  of  things  always  consid- 


100  MECHANICAL  POWERS.  ^  103. 

II.     The  power  ;  long  arm,  and  weight  being  given,  to  find 
the  short  arm. 


_ 


III.     The  iveight,  long  arm,  and  short  arm  being  given,  to 
find  the  power. 


LA 

IV.     The  weight,  short  arm,  and  power  being  given,  to  find 
the  long  arm. 


P 

NOTE.  The  above  formulas  are  equally  applicable  to  either  of  the  three 
kinds  of  levers. 

EXAMPLES  FOR  PRACTICE. 

1.  The  long  arm  of  a  lever  is  70  inches,  the  short  arm  2  inches,  and 
the  weight  900  pounds;  what  power  will  be  required  to  balance  the 
weight  ? 

2.  A  man  has  a  lever  8  ft.  4  in.  long,  resting  upon  a  prop  6  inches 
from  one  end.    If  he  press  upon  the  end  of  the  longer  arm  with  a  force 
of  150  pounds,  what  weight  at  the  end  of  the  shorter  arm  will  be  required 
to  balance  him  ?  Ans.  14100  Ibs. 

3.  A  lever  96  inches  long  has  the  fulcrum  at  one  end,  and  a  power  of 
50  pounds  lifting  at  the  other  ;  what  weight  hung  at  16  inches  from  the 
fulcrum,  will  be  sufficient  to  keep  the  lever  in  a  state  of  rest  ? 

Ans.  300  Ibs. 

4.  A  lever  9  feet  long  is  fastened  at  one  end,  and  has  a  weight  of 
187<5  pounds  at  the  other  ;  how  far  from  the  fulcrum  must  a  power  of 
281<25  pounds  be  applied  to  sustain  the  weight?  Ans.  36|  in. 

5.  What  power  70  inches  from  the  fulcrum,  will  balance  900  pounds  2 
inches  from  the  fulcrum?  Ans.  25f  Ibs. 

6.  The  long  arm  of  a  lever  is  48  inches,  the  power  5  pounds,  and  the 
weight  136  pounds  ;  what  is  the  length  of  the  short  arm  ? 

^F  1O3.  The  ratio  of  the  weight  to  the  power  is  equal  to 
the  ratio  of  the  velocity  of  the  power  to  the  velocity  of  the  weight. 
That  is,  the  power  X  by  the  distance  through  which  it  passes  in 
a  vertical  direction  =  the  weight  X  by  the  distance  through 
which  it  passes  in  a  vertical  direction. 

IT  102.    The  fixed  principle  of  the  lever.    Abbreviations.     1st  formula; 
2d  ;  3d  ;  4th.     Note. 
IT  103.    Principle. 


104,  105. 


MECHANICAL  POWERS.  101 


of  80  pounds  suspended  from  the  end  of  the  long  annJ  ^  ^   ^  ^ 


The  Wheel  and  Axle. 

«IT  1O4  The  Wheel  and  Axle  consists  of  a  wheel  concen- 
tric with  a'cylindrical  axis,  with  which  it  revolves ;  the  power 
being  applied  to  the  circumference  of  the  wheel,  and  the  weight 
to  that  of  the  axis. 


The  wheel  and  axle  is  a 
perpetual  lever,  so  contrived 
as  to  have  a  continued  mo- 
tion about  a  fulcrum.  The 
radius  of  the  wheel  may  be 
regarded  as  the  long  arm  of 
the  lever,  and  the  radius  of 
the  axle  the  short  arm. 
Hence, 


11"  1O5.  A  wheel  and  axle  will  remain  at  rest,  when  the 
ratio  of  the  weight  to  the  power  is  equal  to  the  ratio  of  tie 
radius  of  the  wheel  to  the  radius  of  the  axle.  That  is,  when  the 
weight  X  by  the  radius  of  the  axle  =  the  power  X  by  the 
radius  of  the  wheel.  IT  102.  And, ^ 

IT  1O4      Topic      The  wheel  and  axle.     Its  relation  to  the  lever. 
IT  105.    First  principle;  Second.    Methods  of  applying  power.    Thick- 
ness  of  the  rope.    1st  formula ;  2d  ;  3d ;  4th. 

9* 


102  MECHANICAL  POWERS.  T  105. 

The  ratio  of  the  weight  to  the  power  is  equal  to  the  ratio 
of  the  circumference  of  the  wheel  to  the  circumference  of  the 
axle.  That  is,  the  power  X  by  the  circumference  of  the  wheel 
=  the  weight  X  by  the  circumference  of  the  axle.  IT  103. 

The  power  is  applied  to  the  wheel  and  axle  in  various  ways  : 
sometimes  by  a  rope ;  sometimes  by  pins  which  are  grasped  by 
the  hand,  as  shown  in  the  preceding  diagram ;  and  sometimes 
by  a  winch  or  crank,  as  seen 
in  the  common  windlass. 
But  in  all  cases  the  distance 
from  the  point  at  which  the 
power  is  applied  to  the  cen- 
ter of  the  wheel,  is  to  be 
regarded  as  the  radius  of 
the  wheel.  When  the  thick- 
ness of  the  rope  is  consid- 
ered, the  force  must  be  con- 
ceived as  acting  at  the 
center  of  the  rope,  and  therefore  the  thickness  of  the  rope  must 
be  added  to  the  diameter  of  the  axle ;  and  the  thickness  of  that 
which  supports  the  power,  (if  it  be  applied  by  a  rope  passing 
round  the  wheel,)  must  be  added  to  the  diameter  of  the  wheel. 

The  formulas  given  for  the  lever,  IF  102,  may  be  adapted  to 
the  wheel  and  axle. 

EXAMPLES  FOR  PRACTICE. 

1.  The  diameter  of  the  wheel  is  60  inches,  and  that  of  the  axle  is  7  A 
inches ;  what  weight  upon  the  axle  will  be  balanced  by  a  power  of  24 
pounds  at  the  circumference  of  the  wheel?  A?is.  192  Ibs. 

2.  The  diameter  of  a  wheel  is  35  inches,  and  of  its  axle  9  inches  • 
what  power  applied  at  the  circumference  of  the  wheel,  will  balance  a 
weight  of  2240  pounds  suspended  by  a  rope  1  inch  in  diameter  passing 
around  the  axle?  '  Ans.589^  Ibs. 

3.  The  diameter  of  a  wheel  is  15  feet,  the  weight  is  1000  pounds,  and 
the  power  3  pounds ;  what  must  be  the  diameter  of  the  axle,  that  the 
weight  may  balance  the  power  ? 

4.  The  diameter  of  an  axle  is  8  inches,  the  weight  is  250  pounds,  and 
the  power  20  pounds ;  what  must  be  the  diameter  of  the  wheel,  that  the 
weight  may  balance  the  power  ? 

5.  The  diameter  of  a  wheel  is  8  feet,  and  of  its  axle  20  inches ;  what 
is  the  ratio  of  the  weight  to  the  power?    What  power  will  balance  a 
weight  of  872  pounds,  the  power  and  weight  being  each  sustained  by  a 
rope  1J  inches  in  diameter  ?  ~         ^  to  last    192_ML  lbs> 

6.  The  weight  is  100  pounds,  the  diameter  of  the  axle  15  inches,  and 
the  weight  ascends  1  foot  while  the  power  descends  4 '8  feet :  what  is  the 
diameter  of  the  wheel?    What  is  the  power  ?          Ans  to  last    2Qf  Ibs 


f  106. 


MECHANICAL  POWERS. 


103 


The  Pulley. 

IT  106.  The  Pulley  consists  of  a  wheel,  movable  about 
an  axis,  and  having  a  groove  cut  in  its  circumference,  over 
which  a  cord  passes.  The  wheel  is  generally  called  a  sheave, 
and  is  fixed  in  a  box  called  a  block. 


NOTE. 
tackle. 


A  pulley  is  frequently  called   a 


A  single  pulley  affords  no  mechan- 
ical advantage,  but  serves  merely  to 
change  the  direction  of  the  motion. 
No  mechanical  advantage  is  gained 
from  any  number  of  fixed  pulleys. 


Two  or  more  pulleys,  one  at  least 
of  which  is  movable,  may  be  com- 
bined in  various  ways,  by  which  a 
mechanical  advantage  is  gained, 
greater  or  less,  according  to  the  num- 
ber of  movable  pulleys,  and  the  mode 
of  their  combination.  Thus,  the 
weight  W  is  supported  by  the  two 
parts  of  the  cord  which  passes  under 
the  pulley  E,  J  being  sustained  by 
each  part.  Consequently,  the  power 
P  must  be  J  as  great  as  the  weight 
W,  in  order  to  balance  it. 


VV 


IT  106.    Topic.    Pulley.    Sheave.    Block.    Note.    Use  of  single  pulley . 
Combination  of  pulleys.    Blocks  or  systems  of  pulleys.    Smeaton's  pulley. 


104 


MECHANICAL  POWERS. 


IT  106. 


The  accompanying  diagram  represents  Blocks 
of  Pulleys, .  also  called  a  System  of  Pulleys. 
The  weight  W  is  supported  by  the  6  parts  of 
the  rope  which  passes  under  3  movable,  and 
over  3  fixed  pulleys.  Since  6  parts  of  the  rope 
support  the  whole  weight,  1  part  must  support 
£  of  it.  Therefore,  the  power  P  is  £  as  great 
as  the  weight  W. 


The  combination  of  pulleys  shown  in  this 
diagram,  though  differing  somewhat  from  the 
last  in  form,  is  the  same  in  effect,  consisting  of 
3  movable,  and  3  fixed  pulleys.  But  this  com- 
bination is  liable  to  some  objections,  the  most 
important  of  which  is,  that  unless  the  weight 
is  guided  by  the  hand,  the  ropes  will  twist 
together.  This  objection  led  to  the  invention  of 


W 


107. 


MECHANICAL  POWERS. 


105 


Smeatvrfs  Pulley,  or  Smeatoris  Tack,  as  it  is 
usually  called.  This  contains  2  rows  of  wheels 
in  each  block,  as  shown  in  the  diagram.  A  single 
cord  is  made  to  pass  over  them,  in  such  a  manner, 
that  the  power  and  the  weight  hoth  act  in  the 
same  line  with  the  centers  of  the  2  blocks,  thus 
preventing  the  tendency  to  twist.  The  pupil 
will  readily  trace  the  pulleys,  in  the  order  that  the 
rope  passes  over  them. 


IT  1O7.  A  pulley  will  remain  at  rest,  when  the  power  is  to 
the  weight  as  I  is  to  the  number  of  ropes;  or,  as  1  is  to  twice  We 
number  of  movable  pulleys.  That  is,  the  power  X  by  the  num- 
ber of  ropes,  or  by  twice  the  number  of  movable  pulleys  =  me 
weight.  And,  .  . 

The  ratio  of  the  weight  to  the  power  is  equal  to  the  ratio  of 
the  distance  through  which  the  power  moves  to  the  distance 
through  which  the  weight  moves.  That  is,  the  weight  X  ty  ™e 
distance  through  which  it  moves  =  the  power  X  fy  the  distance 
through  which  it  moves. 

The  formulas  given  for  the  lever,  IT  102,  may  be  adapted  to 
the  pulley. 

EXAMPLES  FOR  PRACTICE. 

1  The  number  of  movable  pulleys  is  4,  and  the  power  is  32  pounds . 
what  weight  will  be  required  to  balance  the  power  ? 

2  The  weight  is  721  pounds,  and  the  number  of  ropes  is  8;  what 
power  will  be  required  to  balance  the  weight  ? 

3  The  weight  is  250  pounds,  and  the  power  12*  pounds  ;  what  is  the 
Dumber  of  movable  pulleys?  What  the  number  rfjgP"^    2Q  ropes- 

4.  The  weight  is  900  pounds,  and  ascends  3  inches 
descends  30  inches;  what  is  the  number  of  ^^ 

IT  107.    First  principle ;  Second.     1st  formula ;  2d  ;  3d  ;  4th. 


106  MECHANICAL  POWERS.  IT  108,  109. 


The  Inclined  Plane. 

IT  1O8.  The  Inclined  Plane  is  a  hard  plane  surface,  form- 
ing an  acute  angle  with  a  horizontal  plane. 

When  a  body  is  moved  upwards  vertically,  its  entire  weight 
must  be  overcome  by  the  power ; 
but  when  it  is  moved  up  an  in- 
clined plane,  a  twofold  effect  is 
produced.  A  part  of  the  weight 
is  sustained  by  the  plane,  and 
the  remainder  presses  against 
any  surface  which  would  resist 
its  motion  down  the  plane.  Thus, 
in  the  accompanying  diagram,  a 
portion  of  the  weight  is  sustained  • 
by  the  plane,  and  the  remainder 
by  the  power. 

If  a  weight  be  moved  along  the 
horizontal  plane  AD,  it  will  be  sup- 
ported by  the  plane ;  if  it  be  moved 
up  the  vertical  plane  DE,  it  will 
be  supported  entirely  by  the  power; 
if  it  be  moved  up  the  inclined  plane 
BE,  at  an  angle  of  45°,  it  will  be 
supported  J  by  the  plane,  and  \  by 
the  power ;  if  it  be  moved  up  the 
plane  CD,  at  an  angle  of  more  than  45°,  less  than  J  of  it  will 
be  supported  by  the  plane,  attd  the  remainder  by  the  power; 
and  if  it  be  moved  up  the  plane  AE,  at  an  angle  of  less  than 
45°,  more  than \  of  it  will  be  supported  by  the  plane,  and  the 
remainder  by  the  power.  And  universally, 

The  power  necessary  to  support  any  given  weight  upon  an 
inclined  plane,  depends  upon  the  length  of  the  plane,  and  the 
abruptness  of  its  ascent. 

IT  1O9.  Hence,  The  power  is  to  the  weight  as  the  hight 
of  the  plane' is  to  its  length.  That  is,  the  'weight  X  by  the 
hight  of  the  plane  =  the  power  X  by  its  length. 

The  formulas  given  for  the  lever,  IF  102,  may  be  adapted  to 
the  inclined  plane. 

IF  108.    Topic.    Inclined  plane.    Analysis.    Different  angles  of  inclined 
planes.    Principle. 
IT  109.    Principle.    1st  formula ;  2d  ;  3d  ;  4th. 


If  HO.  MECHANICAL  POWERS.  107 

EXAMPLES  FOR  PRACTICE. 

1.  The  length  of  an  inclined  plane  is  12  feet,  and  its  hight  is  2  feet ; 
what  power  will  be  necessary  to  support  a  weight  of  200  pounds  upon  the 
plane  ?  Ans.  33J  Ibs. 

2.  A  hill  rises  440  feet  in  half  a  mile ;  what  weight  Mail  a  man,  pull- 
ing with  a  force  of  150  pounds,  be  able  to  keep  from  rolling  down  the 
hill  ?  Ans.  900  Ibs. 

3.  A  boy,  by  bracing  with  a  force  of  70  pounds,  is  able  to  hold  a  bar- 
rel of  oil,  weighing  400  pounds,  upon  an  inclined  plane  15  feet  long ; 
what  is  the  hight  of  the  plane  ?  .      Ans.  2|  ft. 

4.  A  power  of  90  pounds  will  hold  a  weight  of  2700  pounds  upon  an 
inclined  plane  15  feet  high  ;  what  is  the  length  of  the  plane  ?  Ans.  45.0  ft. 


The  Wedge. 

IT  11O.     The  Wedge  consists  of  an  inclined  plane,  or  of 
two  inclined  planes  joined  together,  the  entire  length  of  their 


It  is  sometimes  used  for  raising  bodies,  by  being  made  to 
pass  under  them  ;  but  more  frequently  for  dividing  or  splitting 
them.  In  the  former  case,  if  we  suppose  the  wedge  to  be 
pushed  under  the  load  by  pressure,  its  action  is  precisely  the 
same  as  that  of  the  inclined  plane ;  for  the  effect  is  the  same, 
whether  the  wedge  be  pushed  under  the  load,  or  the  load  be 
drawn  up  the  plane. 

But  the  wedge  is  more  commonly  driven  forward,  by  blows 
from  a  mallet  or  hammer,  while  the  resistances  which  it  has  to 
overcome  act  with  constant  force  against  it.  Hence,  its  power 
cannot  be  estimated  with  any  degree  of  accuracy.  It  may, 
however,  be  stated,  that  the  mechanical  advantage  of  the  wedge 
is  increased  by  diminishing  the  angle  of  its  cutting  edge  ;  but 
tJie  strength  of  the  tool  is  thereby  diminished. 

All  cutting  and  piercing  instruments,  such  as  axes,  knives, 
scissors,  chisels,  &c.,  nails,  pins,  needles,  awls,  &c.,  are  modi- 
fications of  the  wedge.  The  angle  of  the  cutting  edge  of  the 
wedge  is  made  more  or  less  acute,  according  to  the  purpose  to 
which  it  is  to  be  applied.  In  tools  for  cutting  wood,  the  angle 
is  generally  about  30° ;  for  iron,  it  is  from  50°  to  60° ;  and  for 
brass,  from  80°  to  90°.  In  general,  the  softer  the  substance  to 
be  divided  is,  the  more  acute  may  be  the  angle  of  the  wedge ; 
and  tools  which  act  by  pressure  may  have  their  cutting  edges 
more  acute  than  those  which  are  driven  by  a  blow. 

IT  110.  Topic.  Wedge.  Its  uses.  Difficulty  in  estimating  the  power  of 
the  wedge.  Principle.  Examples  of  the  wedge.  Angle  of  the  cutting  edge. 


MECHANICAL  POWERS. 


IT  211. 


The  Screw. 

IT  111*  The  Screw  consists  of  a  spiral  ridge  or  groove, 
winding  round  a  cylinder,  so  as  to  cut  every  line  on  the  surface 
parallel  to  the  axis  at  the  same  angle. 

If  the  inclined  plane 
AC  be  wound  round  a 
cylinder  whose  circum- 
ference equals  the  base 
AB,  the  plane  AC  will 
form  the  thread  of  a 
screw ;  and,  if  the  plane 
be  continued,  the  per- 
pendicular BC  will  be 
the  distance  between  any 
two  contiguous  threads. 
Hence,  The  screw  may  be  supposed  to  consist  of  an  inclined 
plane,  whose  base  is  the  circumference  of  the  screw,  and  whose 
altitude  is  the  perpendicular  distance  between  any  two  con- 
tiguous threads. 

In  the  application  of  the 
screw,  the  weight  is  not  placed 
upon  the  threads,  but  the 
power  is  transmitted  by  caus- 
ing the  screw  to  move  in  a  hol- 
low cylinder,  whose  interior 
surface  contains  a  spiral  cavity 
corresponding  to  the  thread  of 
the  screw,  and  in  which  the 
thread  will  move  by  turning  round  the  screw  continually  in  one 
direction.  This  hollow  cylinder  is  usually  called  the  Nut,  or 
Concave  Screw. 


The  screw  is  seldom  used  alone,  but 
owes  its  efficacy  to  the   lever  with 


Till.    Tbpic.   Screw.    Its  relation  to  the  inclined  plane.    Application  of 
the  screw.  Nut,  Manner  of  applying  the  power.  Movable  screw.  Movable  nut. 


IT  112. 


MECHANICAL  POWERS. 


109 


which  it  is  connected.  The  lever  is 
sometimes  connected  with  the  screw, 
and  sometimes  with  the  nut.  When 
connected  with  the  screw,  the  nut  is 
immovable ;  and  when  with  the  nut, 
the  screw  is  immovable. 


IT  1 12.  A  screw  will  remain  at  rest,  when  the  ratio  of  the 
power  to  the  weight  is  equal  to  the  ratio  of  the  distance  between 
the  adjacent  threads  of  the  screw  to  the  circumference  described 
by  the  point  to  which  the  power  is  applied.  That  is,  the  weight 
X  by  the  distance  between  the  adjacent  threads  =  the  power  X 
by  the  circumference  of  a  circle  whose  radius  is  equal  to  the  length 
of  the  lever. 

Hence,  the  mechanical  advantage  afforded  by  the  screw  is 
dependent  upon  the  fineness  of  the  threads,  the  smallness  of 
the  cylinder  or  body  of  the  screw,  and  the  length  of  the  lever 
by  which  the  power  is  applied. 

The  formulas  given  for  the  lever,  IF  102,  may  be  adapted  to 
the  screw. 

EXAMPLES  FOR  PRACTICE. 

1.  The  threads  of  a  screw  are  1  inch  asunder,  the  length  of  the  lever 
by  which  it  is  turned  is  3£  feet,  and  the  power  is  30  pounds ;  what  weight 
will  be  necessary  to  balance  the  power  ?  Ans.  660  Ibs. 

2.  The  distance  between  the  threads  of  a  screw  is  1£  inches,  and  the 
length  of  the  lever  is  21  inches ;  what  power  is  necessary  to  balance  a 
weight  of  9900  pounds?  Ans.  112<5  Ibs.,  nearly. 

3.  The  weight  is  25000  pounds,  the  power  150  pounds,  and  the  circum- 
ference of  the  circle  described  by  the  power  10  feet ;  what  must  be  the 
distance  between  the  threads  of  the  screw,  that  the  weight  may  balance 
the  power  ?  Ans.  '72  of  an  inch. 

4.  The  threads  of  a  screw  are  |  of  an  inch  apart,  the  weight  is  362057^ 
pounds,  and  the  power  120  pounds  ;  what  must  be  the  length  of  the  lever 
by  which  the  power  is  applied,  that  the  power  will  balance  the  weight? 

Ans.  14'41-f-in. 


IT  112.    Principle. 
2d  ;  3d  ;  4th. 


Mechanical  advantage  of  the  screw.    1st  formula; 


10 


110  MECHANICAL  POWERS.  IT  113— 115. 


Friction. 

IT  113.  Friction  is  the  resistance  produced  by  the  rubbing" 
of  the  surfaces  of  two  solid  bodies  against  each  other. 

If  the  surfaces  of  bodies  were  perfectly  smooth  and  polished, 
they  would  slide  upon  one  another  without  any  resistance  from 
their  contact.  But  this  state  of  smoothness  and  polish  never 
exists,  and  can  never  be  attained.  The  surfaces  of  all  bodies, 
even  when  they  have  received  the  highest  polish  that  we  are 
capable  of  giving  them,  retain  a  greater  or  less  degree  of  rough- 
ness, which  prevents  them  from  sliding  upon  one  another  with- 
out resistance  or  friction. 

Friction  is  of  three  kinds  ;  viz. : 

1st.  That  which  occurs  when  one  body  slides  upon  the  sur- 
face of  another. 

2d.  That  of  rolling  bodies  ;  and, 

3d.  That  of  the  axles  of  wheels. 

V  1141.  From  the  numerous  and  varied  experiments  upon 
this  subject,  have  been  deduced  the  following  conclusions. 

I.     The  friction  of  sliding  lodies. 

1.  Between  similar  substances,  under  similar  circumstances, 
friction  is  a  constant  retarding  force. 

2.  Friction  is  greatest  between  bodies  whose  surfaces  are 
rough,  and  is  lessened  by  polishing  them. 

3.  It  is  greater  between  surfaces  of  the  same  material,  than 
between  those  composed  of  different  materials. 

4.  If  the  rubbing   surfaces   remain  the  same,  the  friction 
increases  directly  as  the  pressure. 

5.  If  the  pressure  remain  the  same,  the  friction  has  no  rela- 
tion to  the  extent  of  the  surface. 

6.  The  application  of  oil,  grease,  or  any  unguent,  in  general 
diminishes  the  friction,  though  in  different  degrees,  dependent 
upon  various  circumstances. 

5T  115.     II.     The  friction  of  rolling  bodies. 
1.  Like  the  friction  of  sliding  bodies,  it  is  a  constant  retard- 
ing force. 

IT  113.  Topic.  Friction.  Causes  of  friction.  1st  kind  of  friction ;  2d 
kind  ;  3d  kind. 

IT  114.  First  law  of  the  friction  of  sliding  bodies  ;  2d  law;  3d  ;  4th; 
5th;  6th. 

IT  115.  First  law  of  the  friction  of  rolling  bodies ;  2d  law;  3d  ;  4th  ; 
5th;  6th. 


1116,117.  MECHANICAL  POWERS.  Ill 

2.  It  is  affected  by  the  nature  of  the  surface,  so  far  as  polish 
is  concerned;  but  is  not  lessened  by  the  application  of  unctuous 
substances. 

3.  It  is   less   between   bodies  of  different   materials,  than 
between  those  of  the  same  substance. 

4.  It  is  directly  proportional  to  the  pressure. 

5.  It  has  no  relation  to  the  extent  of  the  surface, 

6.  It  is  much  less  ia  rolling,  than  in  sliding  bodies, 

IT  1 16.     III.  The  friction  of  the  axles  of  wheels. 

1.  It  is  less  than  that  of  sliding,  but  greater  than  that  of 
rolling  bodies. 

2.  It  fpliows  iri  all  respects  the  laws  of  the  friction  of  sliding 
bodies. 

3.  A  great  advantage  may  be  obtained  from  greasing  the 
surfaces.     By  the  application  of  fresh  tallow,  the*  friction  is 
reduced  about  one  half. 

IT  117.     No  definite  rules  for  the  allowances  which  must 
be  made  for  friction  have  yet  been  established.     The  following 
allowances  are  as  nearly  correct  as  any  that  have  been  pre- 
sented. 
The  friction  of  sliding  bodies  is  frojn  '12 

to  '35  of  the  weight  or  pressure,  or  from  12  to  35  per  cent. 
The  friction  of  rolling  bodies,  • "  5  "  12  " 

"         "       "   the  axles  of  wheels,          "     12  "  20  •    " 
The  friction  of  the  axles  of  wheels  may  be  stated  more  defi- 
nitely, as  follows : 

An  iron  axle  turning  in  a  box  of  oak,    from  12    to  18  per  cent. 
A  wooden  axle  turning  in  a  box  of  v/ood,   "       8    "  15       " 
A  metallic  axle  turning  in  a  box  of  an- 
other  metal,  and  well  coated  with, 
grease,  "      2J  "     6       " 

A  wooden  axle  turning  in  a  box  of  wood, 

and  coated  in  a  similar  manner,  "        3  "    8      " 

An  iron  axle  turning  'in  a  box  of  wood, 

and  coated  in  a  similar  manner,  "        5  "  10      <« 

In  making  allowances  for  friction,  the  per  cent,  of  allowance 
is  to  be  estimated  upon,  and  added  to  the  weight  or  resistance, 
or  estimated  upon,  and  subtracted  from  the  power,  before  any 
other  computations  are  made. 

IT  1 16.    First  l#w  of  the  frictioa  of  the  axles  of  wheels  ;  2d  law  ;  3d. 

IT  1.17.  Rates- per  cent,  of  the  friction  of  sliding  bodies.  Of  rolling  bodies. 
Of  the  axles  of  wheels.  More  definite  statement  of  the  friction  of  the  axles  of 
wheels.  How  allowances  for  friction  are  to  be  made. 


112  MECHANICAL  POWERS.  If  118,  119. 


General  Remarks  upon  the  Mechanical  Powers. 

IT  1 1 8.  The  mechanical  powers  afford  no  positive  gain 
in  any  one  respect,  that  is  not  counterbalanced  by  a  loss  in 
another.  Thus,  one  man,  with  the  aid  of  one  of  the  simple 
mechanical  powers,  is  able  to  remove  a  weight,  or  to  overcome 
a  resistance,  that  would  require  the  united  strength  of  20  men 
unaided  by  any  machine.  Here  is  a  positive  gain  in  one  re- 
spect. But,  to  effect  this  end,  the  one  man  must  cause  the 
power  which  he  applies  to  move  20  times  as  far  as  he  wishes 
to  move  the  weight ;  and  the  time  required  to  perform  the 
labor  will  be  20  times  as  great  as  would  be  required,  had  20 
men  been  employed.  Here,  then,  is  a  loss  in  distance  passed 
over  by  the  one  man,  and  in  the  time  consumed,  fully  equal  to 
the  advantage  afforded  by  the  machine  employed. 

IT  1 19.  The  friction  in  the  several  mechanical  powers  is 
various,  and  is  dependent  upon  the  form  of  the  machine  and 
the  materials  of  which  it  is  composed.  Thus,  the  friction  of  a 
lever,  when  turning  upon  a  sharp-edged  fulcrum  of  hardened 
steel,  is  so  very  small  that,  in  ordinary  cases,  it  is  not  taken 
into  account ;  while,  if  the  fulcrum  be  a  stone  or  a  block  of 
wood,  the  friction  may  be  as  much  as  3  per  cent.  The  friction 
of  the  wheel  and  axle  is  modified  by  various  circumstances, 
which  have  been  named  in  IF  117.  But,  when  cordage  is  em- 
ployed, it  increases  the  resistance  from  7  to  10  per  cent.  In 
the  pulley  the  friction  of  the  cordage,  together  with  that  of  the 
sheaves  and  blocks,  increases  the  resistance  from  20  to  75  per 
cent.  The  friction  of  bodies  rolling  on  inclined  planes,  is  usu- 
ally so  small,  that  in  estimates  it  is  not  considered.  That  of 
sliding  bodies  has  already  been  given, 
11  117.  The  friction  of  the  screw  is 
very  great ;  it  must  exceed  the  resist- 
ance, or  the  screw  will  not  retain  its 
position.  Screws  with  sharp  or  wedge- 
shaped  threads  are  attended  with  more 
friction  than  those  whose  threads  are 
square. 

IT  118.    Topic.    Advantage  of  the  mechanical  powers.    Illustration. 

IT  119.  Causes  of  the  difference  in  the  friction  of  the  different  mechanical 
powers.  Friction  of  the  lever.  Of  the  wheel  and  axle.  Of  the  pulley.  Of 
the  inclined  plane.  Of  the  screw. 


MACHINERY. 


IT  ISO.  Any  machine,  however  simple  or  complex  in  its 
construction  and  operation,  must  contain  one  or  more  of  the 
simple  mechanical  powers  ;  nor  can  it  involve  any  other  prin- 
ciples than  those  of  the  mechanical  powers. 

A  Simple  Machine  is  one  which  involves  but  one  of  the  sim- 
ple mechanical  powers. 

A  Compound  Machine  is  formed  by  combining  two  or  more 
simple  machines. 

IT  131.    Methods  of  Transmitting  Motion. 

Motion  may  be  transmitted  from  the  moving  power  to  the 
other  parts  of  a  machine  in  various  ways,  dependent  upon  cir- 
cumstances-. When  two  parts 
of  a  machine,  acting  at  some 
distance  from  each  other,  are 
to  be  moved  together,  in  the 
same  direction,  the  motion 
may  be  transmitted  by  a  band, 
passing  over  wheels  attached 
to  the  two  parts  of  the  ma- 
chine. And  when  the  two 
parts  to  be  connected  are  to 
move  in  contrary  directions, 
the  band  may  be  crossed.  If 
a  rope  band  be  used,  its  friction,  and  consequently  its  efficacy, 
may  be  increased,  by  grooving  the  edge  of  the  wheels.  And 
when  a  strap  band  is  used,  its  friction  may  be  increased,  by 
increasing  the  width  of  the  band. 


If  133.  But  the  more  common  method  of  transmitting 
motion  is,  by  wheels  having  Teeth  or  Cogs  cut  in  their  circum- 
ference. The  connection  of  toothed  wheels  with  each  other, 

IT  120.  Topic.  Principles  involved  in  every  machine.  Simple  machine. 
Compound  machine. 

IT  121.    Topic.    Bands.    Cross  bands.    Friction  of  bands. 

IT  122.  Gearing.  Pinion.  Its  leaves.  Kinds  of  toothed  wheels.  Spur 
wheel.  Spur  gearing.  Crown  wheel.  Its  effect  when  working  with  a  spur 
wheel.  Bevel  wheel.  Bevel  gearing.  Use  of  bevel  wheels.  Note.  Uni- 
versal joint. 

10* 


114 


MACHINERY 


IT  122. 


for  the  purpose  of  transmitting  motion  in  machinery,  is  called 
Gearing.  It  is  usual  to  call  a  small  wheel  acted  upon  by  a 
large  one,  a  Pinion,  and  its  teeth  the  Leaves  of  the  pinion. 


Toothed  wheels  are  of  three  kinds  ;  viz. : 
1st.  When  the  teeth  are  raised  upon  the 
edge  of  the  wheel,  or  are  perpendicular  to  the 
axis,  the  wheel  is  a  Spur  Wheel ;  and  two 
or  more  spur  wheels  working  together  are 
called  Spur  Gearing. 


2d.  When  the  teeth  are  raised  parallel 
to  the  axis,  or  perpendicular  to  the  plane 
of  the  wheel,  it  is  called  a  Crown  Wheel. 
A  crown  and  a  spur  wheel  working  to- 
gether, serve  to  transmit  the  motion  of 
one  to  the  other  at  a  right  angle. 


3d.  When  the  teeth  are  raised  on  a 
surface  inclined  to  the  plane  of  the  wheel, 
it  is  called  a  Beveled  Wheel ;  and  two  or 
more  beveled  wheels  working  together 
are  called  Beveled  Gearing.  Beveled 
wheels  are  employed  to  transmit  motion 
from  one  axis  to  another  inclined  to  it,  at 
any  proposed  angle. 


NOTE.    Beveled  wheels  are  also  call  *d  conic  il  wheels,  because  their  teeth 
may  be  regarded  as  cut  in  the  frustrum  01  «  ^ 


11  123,  124. 


MACHINERY. 


115 


The  direction  of  motion  may  be  chang- 
ed from  a  right  line  to  any  angle  less 
than  40°,  by  the  Universal  Joint.  This 
is  effected  by  connecting  the  ends  of  two 
axes  with  the  joint,  as  shown  in  the  dia- 
gram. 


IT  123.  Gudgeons,  in  machinery,  are  pins  inserted  in  the 
extremities  of  a  shaft,  or  the  axle  of  a  wheel,  on  which  it 
turns,  and  which  support  the  weight.  In  order  to  diminish 
friction,  gudgeons  are  made  as  small  as  possible  in  diameter ; 
leaving,  however,  sufficient  strength  to  support  the  weight. 

The  Box  of  a  gudgeon  is  the  hollow  cylinder  of  wood  or 
metal,  in  which  the  gudgeon  runs. 


Teeth  of  Wheels. 


IT  124.  Where  the  teeth  of  wheels 
work  into  each  other,  as  represented  in 
the  diagram,  every  part  of  the  side  of 
each  tooth  of  one  wheel  comes  successive- 
ly in  contact  with  a  tooth  of  the  other,  as  the  wheel  turns  round, 
and  consequently  the  force  is  exerted  at  the  points  which 
are  in  contact.  But  it  is  of  the  utmost  importance  that  the 
parts  act  upon  each  other  with  a  uniform  force,  and  with  the 
least  possible  amount  of  friction.  This  end  can  be  attained 


IF  123,    Gudgeons.     Their  size.    Box  of  a  gudgeon. 

IT  124.  Topic.  Object  in  making  the  teeth  of  wheels  curving.  Object  in 
making  the  number  of  teeth  of  two  wheels,  or  of  a  wheel  and  pinion  working 
together,  prime  to  each  other.  Illustration.  Hunting-cog. 


116 


MACHINERY. 


IT  125. 


in  no  other  way  than  by  making  the 
teeth  of  the  wheels  curving.  The  curve 
of  the  teeth  will  be  greater  or  less,  ac- 
cording to  the  size  of  the  wheel  and 
the  dimensions  of  the  teeth. 


But  the  surfaces  of  teeth  will  always  contain  some  inequal- 
ities, and  consequently  will  cause  some  friction.  To  equalize 
the  wear  arising  from  inequalities  on  the  surface  of  the  teeth 
of  wheels  and  pinions,  each  tooth  of  the  pinion  should  work 
in  succession  in  every  tooth  of  the  wheel,  and  not  always  in 
the  same  set  of  teeth.  To  effect  this,  the  number  of  teeth  in 
a  wheel  and  in  a  pinion  which  work  into  each  other,  must  be 
prime  to  each  other.  Thus,  if  the  wheel  contain  61  teeth, 
and  the  pinion  12,  each  tooth  of  the  pinion  will  work  in  suc- 
cession in  every  tooth  of  the  wheel.  In  this  case,  no  tooth  of 
the  pinion  can  act  with  the  same  tooth  of  the  wheel  a  second 
time,  until  it  has  acted  upon  every  other  tooth  of  the  wheel. 
The  odd  tooth  which  produces  this  effect,  is  called,  the  Hunting 
Cog. 


Horse  Power. 

IT  125.  The  force  of  a  machine  or  an  engine  may  be  ob- 
tained and  applied  in  a  variety  of  ways,  as  by  gravity,  animal 
strength,  wind,  water,  steam,  &c. ;  but  in  estimating  the  power 
of  any  machine  of  great  force,  the  power  is  referred  to  a  fixed 
and  established  standard,  called  Horse  Power. 

Horse  Power  is  the  weight  which  a  horse  is  capable  of  rais- 
ing to  a  given  hight  in  a  given  time.  Custom  has  established 
as  a  standard,  that  a  machine  of  one  horse  power  is  capable  of 
raising  a  weight  of  33000  pounds  one  foot  in  a-  minute. 

NOTE.  A  machine  of  1  horse  power  will  raise  a  weight  of  2000  Ibs.  1  rod 
in  a  minute  ;  500  Ibs.  4  rds.  in  a  min. ;  and  125  Ibs.  16  rds.  in  a  min.,  or  3 
miles  an  hour. 

IT  125.  Topic.  Ways  in  which  the  force  of  a  machine  may  be  obtained. 
Fixed  and  established  standard  to  which  the  power  of  any  machine  of  great 
force  is  referred.  Horse  power.  A  machine  of  one-horse  power.  Note. 


126,  127. 


MACHINERY. 


117 


Levers  and  Weighing  Machines. 

IT  12G.  Any  number  of  weights  may  be  attached  to  either 
arm  of  a  lever.  The  lever  will  remain  at  rest,  when  the  sum 
of  tJie  products  of  the  weights  upon  one  arm  X  by  their  respective 
distances  from  the  fulcrum  =  the  sum  of  the  products  of  the 
weights  upon  the  other  arm  X  by  their  respective  distances  from 
the  fulcrum.  See  IT  102. 


Ex.  The  distance  from  E  to  B  is  4  inches,  from  B  to  A  5 
inches,  from  E  to  C  5  inches,  and  from  C  to  D  8  inches  ;  A 
weighs  20  pounds,  B  8  pounds,  and  C  3  pounds.  What  must 
be  the  weight  of  D,  that  the  lever  may  remain  at  rest  ? 

Ans.  15-     Ibs. 


^T  127*.  When  several  simple  levers  act  upon  each  other, 
the  combination  is  called  a  Compound  Lever.  The  principles 
given  for  estimating  the  force  of  a  simple  lever,  IT  102,  are  equal- 
ly applicable  to  the  compound  lever.  But,  by  a  careful  exam- 
ination of  the  operations  necessary  to  estimate  the  force  of  a 
compound  lever,  the  pupil  will  find,  that  a  compound  lever  will 
remain  at  rest,  when  the  product  of  all  the  arms  on  the  side  of 
the  power  X  by  the  power  =  the  product  of  all  the  arms  on  the 
side  of  the  weight  X  by  the  weight. 


A                       C                                      ^    D                        „.    r. 

^  B                               F 
PP 

^^  1 

Ex.  In  the  above  compound  lever  AC  is  9  in.,  BF  12  in., 
DG  10  in.,  BC  2  in.,  DF  2J  in.,  and  EG  2  in. ;  what  weight 
suspended  at  E  will  balance  a  power  of  15  pounds  suspended 
at  A? Ans.  1620  Ibs. 

IT  126.    Topic.    Principle. 

H  127.    Compound  lever.    Principle. 


118  MACHINERY.  IT  128,  129. 


IT  1£8.     The  Balance  consists  of  a  beam  or  lever  suspend- 
ed exactly  in  the  middle,  with  scales  or  basins  hung  at  or 
suspended  from  the  extremities,  of  pre- 
cisely equal  weight.    The  accuracy  of  the 
balance  depends  upon  the  length  of  its 
arms,  and  the  shape  and  material  of  its 
fulcrum. 

A  fraudulent  balance  may  be  made,  by  making  one  arm  of 
the  scale  beam  shorter  than  the  other.  The  fraud  may  readily 
be  detected,  by  weighing  an  article  in  one  scale,  and  then  in 
the  other.  If  it  w.eigh  the  same  in  both,  the  balance  is  correct  ; 
otherwise,  it  -is  fraudulent. 

The  actual  weight  of  a  body  may  be  obtained,  by-  a  false  bal- 
ance, as  follows  : 

1st.  Weigh  the  body  in  the  two  scales  successively. 

2d.  Multiply  the  two  weights  together,  and  extract  the 
square  root  of  their  product; 

NOTE.  The  square  root  of  their  product  is  a  Geometrical  Mean  between 
the  two  weights.  See  Revised  Arjth.,  IT  185,  note. 

Ex.  A  body,  when  placed  in  one  scale  of  a  balance,  weighs 
8£  pounds;  but  when  placed  in  the  other,  it  weighs  14  pounds  ; 
What  is  its  true  weight  ?  Ans.  10'908-f-  Ibs. 

IF  129.  The  Steelyard  is  a  balance,  which  consists  of  a 
lever  having  two  unequal  arms  ;  the  weights  of  bodies  being 
determined  by  means  of  a  single  standard  weight. 

The  body  whose  weight  is  to  be 
determined,  is  suspended  from  the 
extremity  of  the  short  arm  ;  and,  in 
weighing,  the  constant  weight  or 
Counterpoise  (commonly  called  poise,} 
is  moved  along  the  longer  arm,  until 
the  lever  is  brought  to  rest  in  a  hori- 
zontal position.  Divisions  marked  on 
the  longer  arm,  indicate  the  weight  of  any  body  suspended 
from  the  shorter  arm,  balancing  the  poise  at  any  division. 

Ex.  I  have  a  steelyard,  a  weight  of  10  pounds,  and  a  poise 
of  £  pound.  How  will  I  proceed  to  lay  off  pound  notches 
upon  the  long  arm  of  the  steelyard  ? 

IT  128.  Balance.  Its  accuracy.  A  fraudulent  balance.  Manner  of  de- 

tecting it.  How  to  find  the  actual  weight  of  a  body,  by  a  fraudulent  balance. 
Note. 

IT  129.  Steelyard.    Manner  of  using  it. 


1f  130, 131. 


MACHINERY. 


119 


^T  1 3O,  When  the  power  and  the 
weight  do  not  act  on  the  lever  in  di- 
rections perpendicular  to  its  length,  or 
when  the  lever  is  bent  or  crooked,  the 
perpendicular  distances  from  the  ful- 
crum to  the  lines  of  direction  in  which 
the  power  and  weight  act,  are  to  be 
regarded  as  the  arms  of  the  lever. 
Thus,  in  the  present  position  of  the 
Bent  Lever  Balance  represented  in  the 
diagram,  BD  and  BK  are  to  be  con- 
sidered as  the  arms  of  the  bent  lever  CBK.  The  pupil  will 
perceive  that  a  small  weight  in  the  scale  will  elevate  the  weight 
C  but  a  little  distance  upon  the  graduated  scale  FG.  But 
weights  may  be  added,  till  C  shall  be  elevated  to  G.  Every 
change  of  weight  changes  the  relative  distances  of  the  power 
and  weight  from  the  fulcrum. 


Wheel  Work. 

IT  131.  The  Capstan  is  a  strong 
massive  piece  of  timber,  in  the  form 
of  a  cylinder  or  frustrum  of  a  cone, 
around  which  a  rope  is  coiled ;  and 
being  turned  by  means  of  bars  or 
levers,  inserted  into  its  head,  or  a 
drum  attached  to  its  head,  it  affords 
an  advantageous  mode  of  applying 
power  to  overcome  resistance.  The  capstan  is  chiefly  em- 
ployed in  ships  for  weighing  anchors,  hoisting  sails,  &c. ;  and 
on  land,  for  moving  buildings,  &c.  When  used  for  the  last- 
named  purpose,  it  is  commonly  moved  by  horse  power. 

The  power  of  the  capstan  may  be  greatly  increased,  by  con- 
necting with  it  an  arrangement  of  wheel  work. 

PRACTICAL  EXAMPLES  IX  WHEEL  WORK. 

1.  A  capstan  is  1  foot  in  diameter,  the  levers  by  which  it  is  turned  are 
each  6  feet  long,  and  the  rope  to  which  the  weight  is  attached  is  3  inches  in 
diameter ;  allowing  10  per  cent,  for  the  friction  of  the  capstan,  and  1£  per 
cent,  for  the  stiffness  of  the  rope,  what  power  must  be  applied  by  each  of 
5  men,  at  the  end  of  the  levers,  to  move  a  weight  of  12000  pounds  ? 

Ans.  278<75  Ibs. 

IT  130.    Principle  of  the  bent  lever.    Illustration  by  the  bent  lever  balance. 
II  13 !•    Topic.    Capstan.    Where  employed.    Manner  of  increasing  its 
power. 


120 


MACHINERY. 


1T131. 


2.  The  lever  of  a  capstan  2  feet  in  diameter  is  12  feet  long,  and  the 
rope  by  which  the  weight  is  moved  is  2  inches  in  diameter ;  allowing  12 
per  cent,  for  the  friction  of  the  capstan,  and  1  percent,  for  the  stiffness  of 
the  rope,  what  weight  will  be  moved  by  a  horse  attached  to  the  end  of 
the  lever,  and  pulling  with  a  force  of  900  pounds  ?          Ans.  8673T^-  Ibs. 

3.  In  the  spur  gearing 
represented  in  the  diagram, 
the  respective  diameters  of 
the  wheels  A,  B,  and  C, 
are  14,  16,  and  18  inches ; 
and  the  diameters  of  their 
pinions  a,  b,  and  c,  are  3, 
4,  and  5  inches ;  allowing 
5  per  cent,  for  the  friction 
of  the  axles,  and  3  per  cent, 
for  that  of  the  teeth  of  the 
wheels  and  pinions,  what 
power    applied  at   P  will 
be    required    to   move    a 
weight  of  2000  pounds  sus- 
pended at  W  ? 

Ans.  32|  Ibs 

4.  The    respective   circumfer- 
ences of  the  wheels  A,  B,  C,  D, 
and  E,  are  30,  22,  30,  35.  and  44 
inches ;  and  of  their  pinions  a,  b, 
c,  d,  and  e,  10,  10,  10,  11,  and  12 
inches ;    through  what   distance 
•will  the  power  P  move,  while  the 
weight  W  moves  1  foot? 


Ans.  231  feet. 


5.  The  wheels  B,  and  D, 
are  each  10  inches  in  diam- 
eter ;  the  pinions  A,  and  C, 
each  3  inches ;  the  axle  E 
2  inches  ;  the  circumference 
of  the  circle  described  by 
the  power  P  is  33  inches ; 
and  the  rope  which  sustains 
the  weight  is  1  inch  thick. 
Allowing  the  whole  friction 
of  the  machine  to  be  9  per 
cent.,  what  power  applied 
at  P,  will  be  required  to 
raise  a  weight  of  6000 
pounds  suspended  at  W  ? 
Ans. 


1T132. 


MACHINERY. 


6.  A  man  whose  weight  is  150  pounds,  attempts  to  draw 
himself  up,  by  a  rope  passing  over  a  single  fixed  pulley.  Al- 
lowing the  friction  of  the  axle  of  the  pulley  and  of  the  rope 
and  pulley  to  be  20  per  cent.,  with  what  force  must  he  pull 
upon  the  rope  to  effect  his  object  ? 


7.  The  weight  A  is  500  pounds,  the 
friction  of  the  pulley  and  the  rope  at 
B  is  12  per  cent.,  and  of  the  pulley  and 
rope  at  C  13  per  cent. ;  what  strength 
must  the  horse  exert  to  raise  the  weight? 
Ans.  625  Ibs. 


White's  Pulley. 

IT  132.  The  great  amount  of  friction 
offered  by  the  forms  of  pulley  that  have 
been  presented,  renders  their  use  in  some 
measure  objectionable.  The  friction  of  the 
sheaves  and  blocks,  together  with  that  of 
the  cordage  and  the  axles  of  the  sheaves, 
are  sometimes  so  great,  as  to  render  the 
pulley  of  no  advantage.  But  these  objections 
are  removed  in  the  pulley  here  presented, 
and  known  as  White's  Pulley.  The  wheels 
in  each  block  turn  on  the  same  axis,  and 
consequently  revolve  in  the  same  time ; 
and,  instead  of  separate  wheels,  the  upper 
and  lower  blocks  are  each  cut  in  grooves  in 
one  block,  thus  reducing  the  friction  of  the 
sheaves  and  blocks,  and  of  the  axles,  to  that 
of  one  wheel  in  each  block.  The  size  of 
each  wheel  is  so  proportioned  to  the  others, 
that  any  point  in  its  circumference  moves 
with  the  velocity  of  the  rope  on  that  wheel. 
To  effect  this,  the  diameters  of  the  wheels 
in  the  upper  block  must  be  as  the  numbers  1, 
3, 5,  &c.,  and  in  the  lower  block  as  2, 4, 6,  &c. 


IT  133. 


Objections   to  the  common  forms  of  pulley. 


Explanation  of 


122 


MACHINERY. 


IF  133. 


Ex.  The  weight  JV\  in  the  diagram,  is  1200  pounds,  and 
the  resistance  offered  by  the  friction  of  the  pulley  is  15  per 
cent. ;  what  power  applied  at  P,  will  be  necessary  to  raise  the 
weight?  See  IT  107.  Ant.  115  Ibs. 


The  Crane. 

IT  138.  The  Crane  is  a  machine  for  raising  heavy  weights, 
and  depositing  them  at  some  distance  from  their  original  place. 
Its  parts  are  a  jib  or  transverse  beam  CD,  inclined  to  a  perpen- 
dicular in  an  angle  of  40°  or  50°.  This  is  connected  with  the 
vertical  beam  AB,  which  is  fastened  to  the  floor,  but  is  capable 
of  turning  on 
its  axis.  The 
upper  end  of 
the  jib  carries 
a  fixed  pulley 
at  D,  over 
which  passes 
ar ope  or  chain, 
with  a  hook  at 
O  to  support 
the  weight. 
The  wheel- 
work  is  mount- 
ed in  two  cast- 
iron  crosses 
attached  to  the 
beam  AB,  one 
of  which  is 
shown  at  EF- 
GH.  I  is  the 

winch  at  which  the  power  is  applied.  This  carries  a  pinion 
which  works  in  the  wheel  K ;  a  pinion  upon  the  axle  of  the 
wheel  K,  works  in  the  wheel  L ;  and  upon  the  axle  of  the 
wheel  L,  is  a  cylinder  or  barrel,  on  which  the  rope  or  chain 
MNO  is  coiled. 

Ex.  If  the  length  of  the  winch  be  18  in. ;  the  diameter  of  the 
wheel  K  10  in. ;  of  the  wheel  L  24  in. ;  of  the  pinion  attached 
to  the  winch  4  in. ;  of  that  upon  the  axle  of  the  wheel  K  4 

White's  pulley.  Its  advantages  over  the  other  forms.  Diameters  of  the 
•wheels. 

IT  133.  Crane.  The  parts  of  which  it  is  composed.  Application  of  the 
power.  How  to  increase  the  power. 


IF  134,  135. 


MACHINERY. 


123 


in. ;  and  of  the  barrel  M  8  inches ;  what  force  will  be  exerted  at 
W,  by  a  power  of  500  pounds  applied  at  the  winch  ? 

Ans.  33750  Ibs. 


Hunter's  Screw. 

IF  134.  If  the  power  of  the  screw  be  increased,  by 
diminishing1  the  distance  between 
the  threads,  the  strength  of  the 
threads  will  be  so  diminished,  that  a 
slight  resistance  will  tear  them  from 
the  cylinder.  This  inconvenience  is 
removed  by  Hunter's  Screw,  which 
consists  of  two  screws  upon  the  same 
cylinder,  the  threads  being  of  unequal 
fineness.  The  threads  may  have 
any  strength  and  magnitude,  the  effi- 
cacy of  the  screw  depending  not  upon 
the  size  of  the  threads,  but  upon  the 
difference  between  the  distances  of 
the  threads  of  the  two  screws. 

Ex.  The  screw  A  contains  15  threads  to  the  inch,  and  the 
screw  B  16  ;  how  far  will  the  board  D  be  depressed  by  one 
revolution  of  the  screw?  If  the  lever  which  passes  through 
the  head  of  the  screw  A,  be  21  inches  long,  and  it  be  turned 
by  a  power  of  50  pounds,  what  power  will  be  exerted  upon  the 
board  D,  making  a  deduction  of  52  per  cent,  from  the  power 
for  friction?  Ans.  380160  Ibs. 


The  Endless  Screw. 


f  1 35.  The  Endless  Screw  con- 
sists of  a  screw  combined  with  a 
wheel  and  axle  in  such  a  manner, 
that  the  threads  of  the  screw  work 
with  the  teeth  of  the  wheel. 


Ex.  The  winch  is  14  inches  long,  the  threads  of  the  screw 
are  £  inch  apart,  the  radius  of  the  wheel  is  12  inches,  and  of 

IT  134.     Manner  of  increasing  the  power  of  the  screw.     The  result  of 
diminishing  the  distance  between  the  threads.    Hunter's  screw. 
II  135.    Endless  screw. 


124 


MACHINERY. 


1T136. 


the  axle  4  inches ;  what  is  the  ratio  of  the  weight  to  the  power  ? 
What  power  applied  at  P,  will  be  sufficient  to  balance  a  weight 
of  436  pounds  suspended  at  W  ?  Ans.  Iff  Ibs. 


Pumps. 


^T  136.  A  Pump  is  a  machine  for  raising 
water.  The  Common  or  Suction  Pump,  a  section 
of  which  is  represented  in  the  diagram,  is  the 
one  used  for  common  household  purposes.  AC 
is  a  pipe  of  any  convenient  length,  the  lower 
end  of  which  reaches  below  the  surface  of  the 
water  in  the  well  or  reservoir.  The  part  AB  of 
the  pipe  is  commonly  of  greater  diameter  than 
the  part  CH.  V  is  a  valve  opening  upwards. 
P  is  a  piston  moved  by  the  rod  E.  In  this  pis- 
ton is  also  a  valve  opening  upwards.  To  the 
upper  end  of  the  rod  E,  is  attached  the  end  of 
the  short  arm  of  a  lever,  the  end  of  the  long  arm 
being  the  point  at  which  the  power  is  applied* 


Ex.  A  cubic  foot  of  water  weighs  62£  pounds.  Suppose 
the  piston  P  to  be  at  the  bottom  of  the  pipe  AB ;  the  distance 
from  B  to  A,  12  feet,  to  be  filled  with  water ;  the  diameter  of 
the  pipe  A  B  to  be  5  inches ;  the  long  arm  of  the  lever  or  han- 
dle to  be  30  inches ;  the  short  arm  5  inches ;  and  the  friction 
to  be  10  per  cent,  What  power  applied  at  the  end  of  the  long 
arm  of  the  handle,  will  be  required  to  work  the  pump  ?  If  the 
power  move  through  a  vertical  distance  of  50  inches,  what 
quantity  of  water  will  be  discharged  at  one  stroke  of  the  lever  ? 

A        (  Power,  18  Ibs.  11'94-oz. 
**  ?*  1  Water,  163f  cu.  in. 

IT  136*    Pump.    Explanation  of  the  common  or  suction  pump. 


1T 137, 138. 


MACHINERY. 


125 


The  Hydrostatic  Press. 

IT  1 37.  The  Hydrostatic  Press 
is  a  machine  by  which  an  enor- 
mous force  of  pressure  is  obtained 
through  the  medium  of  water.  It 
consists  of  a  short  and  very  strong 
pump  barrel,  with  a  solid  piston,  C. 
To  this  is  attached  the  rod  D, 
mounted  with  the  crosspiece  E, 
which  is  pushed  upwards  against 
the  thing  to  be  compressed.  AB 
is  a  small  pump,  called  a  Forcing 
Pump,  which  drives  or  forces 
water  into  the  pump  barrel  C,  and 
thus  produces  the  pressure.  If  the  small  pump  have  only  y^ 
as  great  an  area  as  the  large  barrel,  a  pressure  of  1  pound  in 
the  pump  AB,  will  produce  a  pressure  of  100  pounds  at  E. 
But  here,  as  in  all  other  cases  in  machinery,  what  is  gained  in 
force,  is  lost  in  distance.  See  IF  118. 

NOTE.  The  hydrostatic  press  is  also  called  the  Hydraulic  Press,  and 
sometimes,  from  the  name  of  the  inventor  of  its  present  fornv  Bf amah's 
Press. 

Ex.  The"  forcing  pump  AB  is  2  inches  in  diameter,  and 
the  barrel  C  18  inches ;  the  piston  of  the  forcing  pump  is 
worked  by  a  lever  25  inches  long,  the  fulcrum  being  at  one  end ; 
and  the  piston  rod  is  attached  to  the  lever  at  the  distance  of  5 
inches  from  the  fulcrum.  What  pressure  will  be  produced  at  E, 
by  a  power  of  140  pounds  applied  at  F,  no  allowance  being 
made  for  friction  ?  Ans.  56700  Ibs. 


Methods  of  Applying  Power  to  Machinery. 

IT  138 .  It  has  been  remarked,  IT  125,  that  the  power  may 
be  applied  to  a  machine  by  gravity,  animal  strength,  wind, 
water,  steam,  &c. 

The  application  of  gravity  and  of  animal  strength  as  moving 
powers,  have  already  been  made  in  the  previous  estimates 
upon  machinery.  But  the  two  are  not  unfrequently  combined, 
as  will  be  seen  in 

IT  137.    Hydrostatic  press.    Explanation  of  the  parts  of  which  it  consists. 
Illustration  of  its  power.    Note. 
1T  138.    Topic.    Reference  to  IT  126.    Gravity  and  animal  strength. 


126 


MACHINERY. 


1F 139, 140. 


The  Tread-Mill. 

IT  139.  The  Tread-Mill 
consists  of  a  wheel,  usually 
about  5  feet  in  diameter,  and 
16  feet  long.  The  circumfer- 
ence is  furnished  with  24  steps, 
on  which  prisoners  are  made 
to  work.  Several  prisoners 
work  together  upon  the  same  wheel,  each  treading  in  a  separate 
compartment.  It  will  readily  be  seen  that  the  weight  of  several 
persons  upon  the  same  side  of  the  wheel,  will  set  it  in  motion. 
When  the  wheel  is  once  in  motion,  the  prisoner  has  no  alterna- 
tive, but  must  keep  treading.  He  is  assisted  in  a  degree,  how- 
ever, by  a  Hand-rail  before  him,  as  shown  in  the  diagram. 

Ex.  A  wheel  5  feet  in  diameter, is  worked  by  5  men,  whose 
average  weight  is  130  pounds.  The  wheel  is  connected  with 
gearing,  which  moves  a  weight  4  feet  while  any  point  in  the 
circumference  of  the  wheel  moves  1  foot.  If  the  power  be 
uniform,  and  the  wheel  revolve  twice  in  a  minute,  what  is  the 
power  of  the  machine,  making  no  allowance  for  friction  ? 

Ans.  $•}£$•  of  1  horse  power. 

IT  14O.  The  tread-mill  is  sometimes  so  constructed  as  to 
be  moved  by  the  weight  of  animals  walking  on  an  inclined 
plane,  as  shown  in  the  diagram. 


Ex.  The  diameter  of  the  wheel  represented  in  the  diagram 
is  18  feet,  and  of  the  axle  16  inches.  The  rope,  which  coils 
round  the  axle,  runs  over  the  crane,  arid  sustains  the  weight,  is 
2  inches  in  diameter.  If  an  animal  exert  a  force  of  400  pounds 

IT  139.    Tread-mill.    Its  construction  and  use. 


11  141.  MACHINERY.  127 

at  the  circumference  of  the  tread-wheel,  what  weight  may  be 
raised  at  the  end  of  the  rope,  deducting  25  per  cent,  from  the 
power,  for  the  friction  of  the  machine  ?  Ans.  3600  Ibs. 

NOTE.  Wind  has  been  employed  to  some  extent  as  an  agent  or  moving 
power  to  machinery.  It  has  in  most  cases  been  superseded  by  animal,  water, 
and  steam  powers,  and  is  now  seldom  employed. 


Water  Wheels. 

5T  141  A  Water  Wheel  is  a  wheel  turned  by  the  force  of 
running  water.  Water  wheels  are  variously  constructed, 
accordingto  the  circumstances  under  which  they  are  intended 
to  act.  They  are  of  three  kinds,  viz.  : 

1st.  The  Overshot  Wheel  is  one  in  which 
the  water  is  brought  over  the  top  of  the 
wheel,  received  in  buckets,  and  by  its  weight 
causes  the  wheel  to  revolve.  It  is  employed 
chiefly  where  the  stream  affords  but  a  small 
supply  of  water  ;  but  in  no  case  can  it  be 
employed,  unless  the  descent  of  the  stream 
be  somewhat  rapid. 

NOTE  1 .  The  action  of  the  water  as  a  moving  power  is  dependent  upon  the 
principles  given  in  TT 130. 

2d.  The  Undershot  Wheel  is  one  in 
which  the  water  strikes  the  float  boards 
below  the  axle.  Its  power  depends  upon 
the  size  of  the  wheel,  and  the  size  and 
velocity  of  the  stream.  It  is  employed 
chiefly  where  the  supply  of  water  is 
abundant,  but  the  banks  of  the  stream 
do  not  admit  of  a  dam  of  sufficient  hight  to  employ  the  over- 
shot wheel  to  advantage. 

3d.  The  Breast  Wheel  is  a  com- 
bination of  the  overshot  and  the 
undershot  wheel,  since  its  force  is 
obtained  partly  from  the  weight,  and 
partly  from  the  velocity  of  the  water. 
It  is  employed  chiefly  where  the  sup- 
ply of  water  is  not  sufficient  to  give  the  required  force  to  an 

IT  141.  Water  wheel.  The  overshot  wheel.  Where  employed.  Note  1. 
The  undershot  wheel.  Where  employed.  The  breast  wheel.  Where  em- 
ployed. Note  2. 


128  MACHINERY.  IF  141. 

undershot  wheel,  and  the  banks  of  the  stream  do  not  admit  of 
a  dam  of  sufficient  hight  to  employ  an  overshot  wheel. 

NOTE  2.  With  a  given  stream  of  water,  under  favorable  circumstances,  the 
power  of  the  overshot  wheel  is  about  twice  that  of  the  breast  wheel ;  and  the 
power  of  the  breast  wheel  from  2  to  5  times  that  of  the  undershot  wheel. 

NOTE  3.  For  an  explanation  of  the  methods  of  generating  steam  power, 
and  applying  it  to  machinery,  the  pupil  is  referred  to  Dr.  Lardner's  Lectures 
on  Science,  Literature,  and  Art ;  and  to  any  of  the  various  treatises  on  Natural 
Philosophy.  And  as  a  study  in  mechanics  and  machinery,  the  locomotive  and 
stationary  steam  engines  furnish  the  pupil  with  a  greater  variety  of  combina- 
tions and  applications  of  the  mechanical  powers,  than  almost  any  other  ma- 
chines. 


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From  the  pen  of  a  distinguished  Clergyman  (Albert  Barnes.) 

11  This  is  the  title  of  a  book  (Rudiments,  &c.)  which  has  evidently  been 
prepared  with  much  care,  and  which  is  intended  to  be  adapted  to  promote 
a  very  important  object  in  schools  and  academies.  Professor  Olmsted  has 
prepared  on  the  same  general  subject,  a  Treatise  on  Natural  Philosophy,  in 
8vo.,  a  Treatise  on  Astronomy,  in  one  vol.  8vo,  a  School  Philosophy,  and  a 
School  Astronomy,  which  have  been  received  with  great  favor  by  the  pub- 
lic, and  which  have  passed  through  numerous  editions.  The  little  work, 
whose  title  is  given  above,  completes  his  plan,  by  adapting  this  kind  of  in- 
struction to  primary  schools.  He  has  some  rare  qualifications  for  the  com- 
position of  such  a  work.  Besides  his  eminence  in  this  department  of 
instruction  in  Yale  College,  and  his  entire  familiarity  with  his  subject,  he 
has  had  the  advantage  of  having  been  himself  a  teacher  in  a  common  school 
and  in  an  academy,  and  of  thus  becoming  acquainted  with  the  best  method 
of  approaching  the  youthful  mind.  The  work  is  prepared  with  great  care, 
with  eminent  ability  and  judgment,  and  is  well  adapted  to  interest  the  class 
of  youth  for  whom  it  is  intended. 

"  The  writer  of  this  notice,  in  commending  this  work  to  the  favorable  re- 
gards of  the  public,  cannot  but  be  struck  with  the  difference  between  such  a 
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